Questions and Answers (Board Exam)

Introduction to Board Exam PYQs

This section contains 30 important board-style questions based on the chapter on Integration. Board examinations reward a clear, stepwise method. For indefinite integrals, always include the constant of integration $C$. For definite integrals, clearly mention the property or theorem being used, such as the substitution rule for definite integrals, symmetry properties, or $$\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx.$$ Showing the method explicitly is often just as important as getting the final answer.

Question 1 [CBSE 2026]

Evaluate: $$\int \frac{x}{(x-1)(x-2)}\,dx$$

Solution: Step 1: Resolve the rational expression into partial fractions: $$\frac{x}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}.$$

Step 2: Multiply both sides by $(x-1)(x-2)$: $$x=A(x-2)+B(x-1).$$

Step 3: Find $A$ and $B$ by suitable values of $x$.

• Put $x=1$: $$1=A(1-2)= -A \implies A=-1.$$
• Put $x=2$: $$2=B(2-1)=B \implies B=2.$$

Step 4: Substitute back: $$\int \frac{x}{(x-1)(x-2)}dx = \int \left(-\frac{1}{x-1}+\frac{2}{x-2}\right)dx.$$

Step 5: Integrate term by term: $$= -\ln|x-1|+2\ln|x-2|+C.$$

Step 6: Combine logarithms: $$= \ln|(x-2)^2| - \ln|x-1| + C = \ln\left|\frac{(x-2)^2}{x-1}\right| + C.$$

Answer: $$\ln\left|\frac{(x-2)^2}{x-1}\right| + C$$

Question 2 [CBSE 2024]

Evaluate: $$\int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx$$

Solution: Step 1: Let $$I=\int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx \quad ...(1)$$

Step 2: Use the property $$\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx,$$ with $a=\pi/2$: $$I=\int_0^{\pi/2} \frac{\sqrt{\sin(\pi/2-x)}}{\sqrt{\sin(\pi/2-x)}+\sqrt{\cos(\pi/2-x)}}\,dx.$$ Using complementary angle identities, $$\sin(\pi/2-x)=\cos x, \qquad \cos(\pi/2-x)=\sin x,$$ so $$I=\int_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx \quad ...(2)$$

Step 3: Add (1) and (2): $$2I=\int_0^{\pi/2} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx = \int_0^{\pi/2}1\,dx.$$

Step 4: Evaluate: $$2I=[x]_0^{\pi/2}=\frac{\pi}{2}.$$ Hence, $$I=\frac{\pi}{4}.$$

Answer: $$\frac{\pi}{4}$$

Question 3 [CBSE 2025]

Evaluate: $$\int e^x\left(\frac{1+\sin x}{1+\cos x}\right)dx$$

Solution: Step 1: Rewrite the trigonometric fraction in a useful form. Observe that $$\frac{1+\sin x}{1+\cos x}=\frac{1}{1+\cos x}+\frac{\sin x}{1+\cos x}.$$ Now, $$\frac{1}{1+\cos x}=\frac12\sec^2\left(\frac{x}{2}\right)$$ using $1+\cos x=2\cos^2(x/2)$, and $$\frac{\sin x}{1+\cos x}=\tan\left(\frac{x}{2}\right)$$ using $\sin x=2\sin(x/2)\cos(x/2)$. Therefore, $$\frac{1+\sin x}{1+\cos x}=\tan\left(\frac{x}{2}\right)+\frac12\sec^2\left(\frac{x}{2}\right).$$

Step 2: Let $$f(x)=\tan\left(\frac{x}{2}\right).$$ Then $$f'(x)=\frac12\sec^2\left(\frac{x}{2}\right).$$ So the integrand becomes $$e^x[f(x)+f'(x)].$$

Step 3: Use the standard result $$\int e^x[f(x)+f'(x)]dx=e^x f(x)+C.$$ Thus, $$\int e^x\left(\frac{1+\sin x}{1+\cos x}\right)dx = e^x\tan\left(\frac{x}{2}\right)+C.$$

Answer: $$e^x\tan\left(\frac{x}{2}\right) + C$$

Question 4 [CBSE 2023]

Evaluate: $$\int \frac{dx}{\sqrt{9+8x-x^2}}$$

Solution: Step 1: Complete the square in the expression under the root: $$9+8x-x^2 = -(x^2-8x-9).$$ A cleaner way is: $$9+8x-x^2 = 9-(x^2-8x).$$ Now add and subtract $16$ inside the bracket: $$=9-(x^2-8x+16-16)=9-(x-4)^2+16=25-(x-4)^2.$$ Thus, $$9+8x-x^2 = 5^2-(x-4)^2.$$

Step 2: The integral becomes $$\int \frac{dx}{\sqrt{5^2-(x-4)^2}}.$$

Step 3: Use the standard formula $$\int \frac{dx}{\sqrt{a^2-u^2}} = \sin^{-1}\left(\frac{u}{a}\right)+C,$$ where here $u=x-4$ and $a=5$.

Therefore, $$\int \frac{dx}{\sqrt{9+8x-x^2}} = \sin^{-1}\left(\frac{x-4}{5}\right)+C.$$

Answer: $$\sin^{-1}\left(\frac{x-4}{5}\right) + C$$

Question 5 [CBSE 2026]

Evaluate: $$\int_0^\pi \frac{x\tan x}{\sec x+\tan x}\,dx$$

Solution: Step 1: Simplify the integrand: $$\frac{\tan x}{\sec x+\tan x} = \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+\frac{\sin x}{\cos x}} = \frac{\sin x}{1+\sin x}.$$ So, $$I=\int_0^\pi \frac{x\sin x}{1+\sin x}\,dx \quad ...(1)$$

Step 2: Use the property $$\int_0^a f(x)dx = \int_0^a f(a-x)dx$$ with $a=\pi$: $$I=\int_0^\pi \frac{(\pi-x)\sin(\pi-x)}{1+\sin(\pi-x)}dx.$$ Since $\sin(\pi-x)=\sin x$, $$I=\int_0^\pi \frac{(\pi-x)\sin x}{1+\sin x}dx \quad ...(2)$$

Step 3: Add (1) and (2): $$2I=\int_0^\pi \frac{\pi\sin x}{1+\sin x}dx = \pi\int_0^\pi \frac{\sin x}{1+\sin x}dx.$$

Step 4: Rewrite the fraction: $$\frac{\sin x}{1+\sin x}=\frac{1+\sin x-1}{1+\sin x}=1-\frac{1}{1+\sin x}.$$ Hence, $$2I=\pi\int_0^\pi \left(1-\frac{1}{1+\sin x}\right)dx.$$

Step 5: Rationalize the second fraction: $$\frac{1}{1+\sin x} = \frac{1-\sin x}{1-\sin^2 x}=\frac{1-\sin x}{\cos^2 x}=\sec^2 x-\sec x\tan x.$$ Therefore, $$1-\frac{1}{1+\sin x}=1-\sec^2 x+\sec x\tan x.$$ So, $$2I=\pi\int_0^\pi (1-\sec^2 x+\sec x\tan x)dx.$$

Step 6: Integrate: $$\int (1-\sec^2 x+\sec x\tan x)dx = x-\tan x+\sec x.$$ Thus, $$2I=\pi[x-\tan x+\sec x]_0^\pi.$$

Step 7: Evaluate the limits: At $x=\pi$: $$\pi-\tan\pi+\sec\pi = \pi-0-1=\pi-1.$$ At $x=0$: $$0-\tan 0+\sec 0=1.$$ Therefore, $$2I=\pi[(\pi-1)-1]=\pi(\pi-2).$$ Hence, $$I=\frac{\pi}{2}(\pi-2).$$

Answer: $$\frac{\pi}{2}(\pi-2)$$

Question 6 [CBSE 2024]

Evaluate: $$\int x\log x\,dx$$

Solution: Step 1: Use integration by parts. Choose $$u=\log x, \qquad dv=x\,dx$$ according to the ILATE rule.

Step 2: Differentiate and integrate: $$du=\frac{1}{x}dx, \qquad v=\frac{x^2}{2}.$$

Step 3: Apply the formula $$\int u\,dv = uv - \int v\,du.$$ So, $$\int x\log x\,dx = \log x\cdot \frac{x^2}{2} - \int \frac{x^2}{2}\cdot \frac{1}{x}dx.$$

Step 4: Simplify the second integral: $$= \frac{x^2}{2}\log x - \frac12\int x\,dx$$ $$= \frac{x^2}{2}\log x - \frac12\cdot \frac{x^2}{2} + C$$ $$= \frac{x^2}{2}\log x - \frac{x^2}{4} + C.$$

Answer: $$\frac{x^2}{2}\log x - \frac{x^2}{4} + C$$

Question 7 [CBSE 2025]

Evaluate: $$\int_{-1}^1 \log\left(\frac{2-x}{2+x}\right)dx$$

Solution: Step 1: Let $$f(x)=\log\left(\frac{2-x}{2+x}\right).$$ Since the limits are symmetric, check whether $f(x)$ is even or odd.

Step 2: Compute $f(-x)$: $$f(-x)=\log\left(\frac{2+x}{2-x}\right).$$ Using $$\log\left(\frac{a}{b}\right)=-\log\left(\frac{b}{a}\right),$$ we get $$f(-x)=-\log\left(\frac{2-x}{2+x}\right)=-f(x).$$ So $f(x)$ is an odd function.

Step 3: The integral of an odd function over $[-a,a]$ is zero: $$\int_{-1}^1 f(x)dx=0.$$

Answer: $$0$$

Question 8 [CBSE 2022]

Evaluate: $$\int \frac{2x+1}{\sqrt{x^2+2x-1}}\,dx$$

Solution: Step 1: The derivative of the expression inside the square root is $$\frac{d}{dx}(x^2+2x-1)=2x+2,$$ which is close to the numerator $2x+1$. So write: $$2x+1=(2x+2)-1.$$ Hence, $$I=\int \frac{2x+2}{\sqrt{x^2+2x-1}}dx - \int \frac{1}{\sqrt{x^2+2x-1}}dx.$$ Let these be $I_1-I_2$.

Step 2: Evaluate $$I_1=\int \frac{2x+2}{\sqrt{x^2+2x-1}}dx.$$ Put $$t=x^2+2x-1 \implies dt=(2x+2)dx.$$ Then $$I_1=\int t^{-1/2}dt=2\sqrt{t}=2\sqrt{x^2+2x-1}.$$

Step 3: Evaluate $$I_2=\int \frac{1}{\sqrt{x^2+2x-1}}dx.$$ Complete the square: $$x^2+2x-1=(x+1)^2-2=(x+1)^2-(\sqrt2)^2.$$ Now use the standard formula $$\int \frac{dx}{\sqrt{u^2-a^2}} = \ln|u+\sqrt{u^2-a^2}|+C,$$ with $u=x+1$ and $a=\sqrt2$: $$I_2=\ln|x+1+\sqrt{x^2+2x-1}|.$$

Step 4: Combine: $$I=2\sqrt{x^2+2x-1}-\ln|x+1+\sqrt{x^2+2x-1}|+C.$$

Answer: $$2\sqrt{x^2+2x-1} - \log|x+1+\sqrt{x^2+2x-1}| + C$$

Question 9 [CBSE 2026]

Evaluate: $$\int_1^3 |x^2-4|\,dx$$

Solution: Step 1: Find where the expression inside the modulus becomes zero: $$x^2-4=0 \implies x=\pm 2.$$ In the interval $[1,3]$, only $x=2$ lies inside.

Step 2: Split the integral at $x=2$: $$I=\int_1^2 |x^2-4|dx + \int_2^3 |x^2-4|dx.$$

Step 3: Determine the sign in each interval:

• For $1<x<2$, $x^2<4$, so $x^2-4<0$ and $$|x^2-4|=4-x^2.$$
• For $2<x<3$, $x^2>4$, so $$|x^2-4|=x^2-4.$$

Step 4: Rewrite: $$I=\int_1^2 (4-x^2)dx + \int_2^3 (x^2-4)dx.$$

Step 5: Integrate: $$\int (4-x^2)dx = 4x-\frac{x^3}{3}, \qquad \int (x^2-4)dx = \frac{x^3}{3}-4x.$$ So, $$I=\left[4x-\frac{x^3}{3}\right]_1^2 + \left[\frac{x^3}{3}-4x\right]_2^3.$$

Step 6: Evaluate first part: $$\left(8-\frac{8}{3}\right)-\left(4-\frac13\right)=\frac{16}{3}-\frac{11}{3}=\frac53.$$ Second part: $$\left(9-12\right)-\left(\frac83-8\right)= -3 - \left(-\frac{16}{3}\right)=\frac73.$$ Thus, $$I=\frac53+\frac73=\frac{12}{3}=4.$$

Answer: $$4$$

Question 10 [CBSE 2023]

Evaluate: $$\int \frac{dx}{x(x^n+1)}$$

Solution: Step 1: Multiply numerator and denominator by $x^{n-1}$: $$\int \frac{x^{n-1}}{x^n(x^n+1)}dx.$$

Step 2: Let $$t=x^n+1.$$ Then $$dt=nx^{n-1}dx \implies x^{n-1}dx=\frac{dt}{n}.$$ Also, $$x^n=t-1.$$

Step 3: Substitute: $$\int \frac{1}{x(x^n+1)}dx = \frac1n \int \frac{dt}{t(t-1)}.$$

Step 4: Resolve into partial fractions: $$\frac{1}{t(t-1)}=\frac{1}{t-1}-\frac{1}{t}.$$ So, $$\frac1n\int \left(\frac{1}{t-1}-\frac{1}{t}\right)dt.$$

Step 5: Integrate: $$= \frac1n\left(\ln|t-1| - \ln|t|\right)+C = \frac1n\ln\left|\frac{t-1}{t}\right|+C.$$

Step 6: Put back $t=x^n+1$: $$= \frac1n\ln\left|\frac{x^n}{x^n+1}\right|+C.$$

Answer: $$\frac{1}{n}\log\left|\frac{x^n}{x^n+1}\right| + C$$

Question 11 [CBSE 2025]

Evaluate: $$\int_0^{\pi/2} \log(\sin x)dx$$

Solution: Step 1: Let $$I=\int_0^{\pi/2} \log(\sin x)dx \quad ...(1)$$

Step 2: Use the property $$\int_0^a f(x)dx=\int_0^a f(a-x)dx$$ with $a=\pi/2$: $$I=\int_0^{\pi/2} \log(\sin(\pi/2-x))dx=\int_0^{\pi/2} \log(\cos x)dx \quad ...(2)$$

Step 3: Add (1) and (2): $$2I=\int_0^{\pi/2} [\log(\sin x)+\log(\cos x)]dx = \int_0^{\pi/2} \log(\sin x\cos x)dx.$$ Now, $$\sin x\cos x = \frac{\sin 2x}{2}.$$ So, $$2I=\int_0^{\pi/2} \log\left(\frac{\sin 2x}{2}\right)dx = \int_0^{\pi/2} \log(\sin 2x)dx - \int_0^{\pi/2}\log 2\,dx.$$

Step 4: Let $$I_1=\int_0^{\pi/2}\log(\sin 2x)dx.$$ Put $2x=t$, so $dt=2dx$, $dx=dt/2$. Limits change from $0$ to $\pi$: $$I_1=\frac12\int_0^{\pi}\log(\sin t)dt.$$ Now use symmetry about $\pi/2$: $$\int_0^{\pi}\log(\sin t)dt = 2\int_0^{\pi/2}\log(\sin t)dt = 2I.$$ Hence, $$I_1=I.$$

Step 5: Therefore, $$2I=I-\left[ x\log 2 \right]_0^{\pi/2}=I-\frac{\pi}{2}\log 2.$$ So, $$I=-\frac{\pi}{2}\log 2.$$

Answer: $$-\frac{\pi}{2}\log 2$$

Question 12 [CBSE 2024]

Evaluate: $$\int \frac{\cos x}{(1-\sin x)(2-\sin x)}dx$$

Solution: Step 1: Let $$t=\sin x \implies dt=\cos x\,dx.$$ Then the integral becomes $$\int \frac{dt}{(1-t)(2-t)}.$$

Step 2: Resolve into partial fractions: $$\frac{1}{(1-t)(2-t)} = \frac{A}{1-t}+\frac{B}{2-t}.$$ Multiplying through: $$1=A(2-t)+B(1-t).$$

Step 3: Find $A$ and $B$:

• Put $t=1$: $$1=A \implies A=1.$$
• Put $t=2$: $$1=-B \implies B=-1.$$

Step 4: So, $$\int \left(\frac{1}{1-t}-\frac{1}{2-t}\right)dt.$$ Now, $$\int \frac{1}{1-t}dt = -\ln|1-t|, \qquad \int \frac{1}{2-t}dt = -\ln|2-t|.$$ Thus, $$= -\ln|1-t| + \ln|2-t| + C = \ln\left|\frac{2-t}{1-t}\right| + C.$$

Step 5: Substitute back $t=\sin x$.

Answer: $$\log\left|\frac{2-\sin x}{1-\sin x}\right| + C$$

Question 13 [CBSE 2026]

Evaluate: $$\int \frac{x^2}{(x^2+1)(x^2+4)}dx$$

Solution: Step 1: Since the denominator is a product of quadratic factors, try decomposition of the form $$\frac{x^2}{(x^2+1)(x^2+4)} = \frac{A}{x^2+1}+\frac{B}{x^2+4}.$$

Step 2: Multiply both sides by $(x^2+1)(x^2+4)$: $$x^2 = A(x^2+4) + B(x^2+1).$$ That is, $$x^2=(A+B)x^2 + (4A+B).$$

Step 3: Compare coefficients: $$A+B=1, \qquad 4A+B=0.$$ Subtracting, $$3A=-1 \implies A=-\frac13.$$ Then, $$B=1-\left(-\frac13\right)=\frac43.$$

Step 4: Rewrite the integral: $$\int \left(-\frac{1}{3(x^2+1)} + \frac{4}{3(x^2+4)}\right)dx.$$

Step 5: Use the standard formula $$\int \frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\left(\frac{x}{a}\right)+C.$$ So, $$= -\frac13\tan^{-1}x + \frac43\cdot \frac12\tan^{-1}\left(\frac{x}{2}\right)+C$$ $$= -\frac13\tan^{-1}x + \frac23\tan^{-1}\left(\frac{x}{2}\right)+C.$$

Answer: $$\frac{2}{3}\tan^{-1}\left(\frac{x}{2}\right) - \frac{1}{3}\tan^{-1}(x) + C$$

Question 14 [CBSE 2022]

Evaluate: $$\int_0^{\pi/4} \frac{\sin x+\cos x}{9+16\sin 2x}dx$$

Solution: Step 1: Let $$t=\sin x-\cos x.$$ Then $$dt=(\cos x+\sin x)dx,$$ which matches the numerator.

Step 2: Express $\sin 2x$ in terms of $t$. Since $$t^2=(\sin x-\cos x)^2=\sin^2 x+\cos^2 x-2\sin x\cos x = 1-\sin 2x,$$ we get $$\sin 2x = 1-t^2.$$

Step 3: Change the limits:

• At $x=0$: $$t=\sin 0 - \cos 0 = -1.$$
• At $x=\pi/4$: $$t=\sin(\pi/4)-\cos(\pi/4)=0.$$

Step 4: Substitute into the integral: $$\int_{-1}^{0} \frac{dt}{9+16(1-t^2)} = \int_{-1}^{0} \frac{dt}{25-16t^2}.$$

Step 5: Factor: $$25-16t^2 = 16\left(\left(\frac54\right)^2 - t^2\right).$$ So, $$= \frac{1}{16}\int_{-1}^{0} \frac{dt}{\left(\frac54\right)^2 - t^2}.$$

Step 6: Use the standard formula $$\int \frac{dx}{a^2-x^2} = \frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right|+C.$$ Here $a=5/4$, so $$= \frac{1}{16}\cdot \frac{1}{2(5/4)}\left[\ln\left|\frac{5/4+t}{5/4-t}\right|\right]_{-1}^{0}$$ $$= \frac{1}{40}\left[\ln\left|\frac{5+4t}{5-4t}\right|\right]_{-1}^{0}.$$

Step 7: Evaluate: At $t=0$, logarithm is $\ln 1=0$. At $t=-1$, it is $$\ln\left(\frac{1}{9}\right).$$ So, $$I=\frac{1}{40}\left(0-\ln\frac19\right)=\frac{1}{40}\ln 9 = \frac{1}{20}\ln 3.$$

Answer: $$\frac{1}{20}\log 3$$

Question 15 [CBSE 2025]

Evaluate: $$\int \frac{e^x(1+x)}{\cos^2(xe^x)}dx$$

Solution: Step 1: Let $$t=xe^x.$$ Differentiate using the product rule: $$dt = (1\cdot e^x + x\cdot e^x)dx = e^x(1+x)dx.$$ This is exactly the numerator.

Step 2: Substitute into the integral: $$\int \frac{dt}{\cos^2 t} = \int \sec^2 t\,dt.$$

Step 3: Integrate: $$\int \sec^2 t\,dt = \tan t + C.$$

Step 4: Replace $t$ by $xe^x$.

Answer: $$\tan(xe^x) + C$$

Question 16 [CBSE 2023]

Evaluate: $$\int_{-1}^1 x^5\sqrt{a^2-x^2}\,dx$$

Solution: Step 1: The limits are symmetric about $0$, so test whether the integrand is even or odd. Let $$f(x)=x^5\sqrt{a^2-x^2}.$$

Step 2: Compute $f(-x)$: $$f(-x)=(-x)^5\sqrt{a^2-(-x)^2}=-x^5\sqrt{a^2-x^2}=-f(x).$$ So $f(x)$ is odd.

Step 3: The definite integral of an odd function over $[-1,1]$ is zero: $$\int_{-1}^1 f(x)dx = 0.$$

Answer: $$0$$

Question 17 [CBSE 2026]

Evaluate: $$\int e^{2x}\sin x\,dx$$

Solution: Step 1: Let $$I=\int e^{2x}\sin x\,dx.$$ Apply integration by parts with $$u=\sin x, \qquad dv=e^{2x}dx.$$ Then $$du=\cos x\,dx, \qquad v=\frac{1}{2}e^{2x}.$$ So, $$I=\frac12 e^{2x}\sin x - \frac12\int e^{2x}\cos x\,dx.$$

Step 2: Let $$J=\int e^{2x}\cos x\,dx.$$ Again apply integration by parts: $$u=\cos x, \qquad dv=e^{2x}dx,$$ so $$du=-\sin x\,dx, \qquad v=\frac12 e^{2x}.$$ Thus, $$J=\frac12 e^{2x}\cos x + \frac12\int e^{2x}\sin x\,dx = \frac12 e^{2x}\cos x + \frac12 I.$$

Step 3: Substitute $J$ into the expression for $I$: $$I=\frac12 e^{2x}\sin x - \frac12\left(\frac12 e^{2x}\cos x + \frac12 I\right).$$ So, $$I=\frac12 e^{2x}\sin x - \frac14 e^{2x}\cos x - \frac14 I.$$

Step 4: Bring $I$ terms together: $$I+\frac14 I = \frac54 I = \frac12 e^{2x}\sin x - \frac14 e^{2x}\cos x.$$ Hence, $$I=\frac45\left(\frac12 e^{2x}\sin x - \frac14 e^{2x}\cos x\right).$$ Simplify: $$I=\frac{e^{2x}}{5}(2\sin x-\cos x)+C.$$

Answer: $$\frac{e^{2x}}{5}(2\sin x-\cos x) + C$$

Question 18 [CBSE 2024]

Evaluate: $$\int_0^{\pi/2} \frac{1}{1+\sqrt{\tan x}}dx$$

Solution: Step 1: Rewrite in terms of sine and cosine: $$\frac{1}{1+\sqrt{\tan x}} = \frac{1}{1+\frac{\sqrt{\sin x}}{\sqrt{\cos x}}} = \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}.$$ Hence, $$I=\int_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx \quad ...(1)$$

Step 2: Use the property $$\int_0^a f(x)dx=\int_0^a f(a-x)dx$$ with $a=\pi/2$: $$I=\int_0^{\pi/2} \frac{\sqrt{\cos(\pi/2-x)}}{\sqrt{\cos(\pi/2-x)}+\sqrt{\sin(\pi/2-x)}}dx.$$ Now, $$\cos(\pi/2-x)=\sin x, \qquad \sin(\pi/2-x)=\cos x,$$ so $$I=\int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx \quad ...(2)$$

Step 3: Add (1) and (2): $$2I=\int_0^{\pi/2}1\,dx=\frac{\pi}{2}.$$ Therefore, $$I=\frac{\pi}{4}.$$

Answer: $$\frac{\pi}{4}$$

Question 19 [CBSE 2025]

Evaluate: $$\int \frac{dx}{\sqrt{x}+x}$$

Solution: Step 1: Put $$x=t^2 \implies dx=2t\,dt, \qquad \sqrt{x}=t.$$

Step 2: Substitute: $$\int \frac{2t\,dt}{t+t^2} = \int \frac{2t}{t(1+t)}dt.$$ Cancel the common factor $t$: $$= \int \frac{2}{1+t}dt.$$

Step 3: Integrate: $$= 2\ln|1+t|+C.$$

Step 4: Put back $t=\sqrt{x}$: $$=2\ln|1+\sqrt{x}|+C.$$ Since for the usual domain $x\ge 0$, we have $1+\sqrt{x}>0$, this is commonly written without modulus.

Answer: $$2\log(1+\sqrt{x}) + C$$

Question 20 [CBSE 2023]

Evaluate: $$\int e^x\left(\frac{1-x}{1+x^2}\right)^2 dx$$

Solution: Step 1: Expand the square in the numerator: $$\left(\frac{1-x}{1+x^2}\right)^2 = \frac{(1-x)^2}{(1+x^2)^2} = \frac{1-2x+x^2}{(1+x^2)^2}.$$ Now split the numerator: $$1-2x+x^2 = (1+x^2) - 2x.$$ Hence, $$\frac{1-2x+x^2}{(1+x^2)^2} = \frac{1+x^2}{(1+x^2)^2} - \frac{2x}{(1+x^2)^2} = \frac{1}{1+x^2} - \frac{2x}{(1+x^2)^2}.$$

Step 2: Let $$f(x)=\frac{1}{1+x^2}.$$ Then $$f'(x)=-\frac{2x}{(1+x^2)^2}.$$ So the integrand is exactly $$e^x[f(x)+f'(x)].$$

Step 3: Use the theorem $$\int e^x[f(x)+f'(x)]dx = e^x f(x)+C.$$ Thus, $$\int e^x\left(\frac{1-x}{1+x^2}\right)^2 dx = \frac{e^x}{1+x^2}+C.$$

Answer: $$\frac{e^x}{1+x^2} + C$$

Question 21 [CBSE 2026]

Evaluate: $$\int_0^1 x(1-x)^5\,dx$$

Solution: Step 1: Use the property $$\int_0^a f(x)dx = \int_0^a f(a-x)dx$$ with $a=1$. Thus, $$\int_0^1 x(1-x)^5dx = \int_0^1 (1-x)x^5dx.$$

Step 2: Expand the simpler integrand: $$= \int_0^1 (x^5-x^6)dx.$$

Step 3: Integrate term by term: $$= \left[\frac{x^6}{6} - \frac{x^7}{7}\right]_0^1.$$

Step 4: Evaluate: $$= \frac16 - \frac17 = \frac{7-6}{42}=\frac{1}{42}.$$

Answer: $$\frac{1}{42}$$

Question 22 [CBSE 2024]

Evaluate: $$\int_0^4 |x-1|dx$$

Solution: Step 1: The expression inside the modulus becomes zero at $$x=1.$$ So split the interval at $x=1$: $$I=\int_0^1 |x-1|dx + \int_1^4 |x-1|dx.$$

Step 2: Decide the sign in each interval:

• For $0\le x<1$, $x-1<0$, so $$|x-1|=1-x.$$
• For $1\le x\le 4$, $x-1\ge 0$, so $$|x-1|=x-1.$$

Step 3: Rewrite: $$I=\int_0^1 (1-x)dx + \int_1^4 (x-1)dx.$$

Step 4: Integrate: $$\int (1-x)dx = x-\frac{x^2}{2}, \qquad \int (x-1)dx = \frac{x^2}{2}-x.$$ So, $$I=\left[x-\frac{x^2}{2}\right]_0^1 + \left[\frac{x^2}{2}-x\right]_1^4.$$

Step 5: Evaluate: First part: $$\left(1-\frac12\right)-0=\frac12.$$ Second part: $$\left(8-4\right)-\left(\frac12-1\right)=4-\left(-\frac12\right)=\frac92.$$ Hence, $$I=\frac12+\frac92=5.$$

Answer: $$5$$