Board Exam-Pattern Questions

How This Bank Works

Note: These are original practice questions written in the CBSE/State Board pattern — matched to the style, difficulty and marking scheme of board questions on Integrals. They are not reproductions of any specific year's paper.

Integrals is the highest-weight calculus chapter in the board paper. The recurring slots:

1. 2-mark: one-step standard integrals, a direct substitution, or a short definite evaluation.
2. 3-mark: by parts, a partial-fractions decomposition, completing the square, or a definite integral with one property.
3. 5-mark: the long definite integral by properties (King's rule + add), or a full partial-fractions/by-parts chain.

Marking-scheme habits that earn full credit:

1. Write the substitution and $dt$ explicitly — a bare answer loses the method mark.
2. In definite integrals by substitution, show the new limits in the same line.
3. Keep $+C$ on every indefinite integral — dropping it costs half a mark.
4. For properties questions, name the property used ("by $\int_0^a f(x)dx = \int_0^a f(a-x)dx$").

Attempt each question before opening its solution.

2-Mark Questions

Q1. Find $\displaystyle\int \sec x\,(\sec x + \tan x)\,dx$.

Solution:

1. Expand: $\int\left(\sec^2 x + \sec x\tan x\right)dx$.
2. Integrate term-wise: both are table forms: $\tan x$ and $\sec x$.

Answer: $\tan x + \sec x + C$.

Q2. Evaluate $\displaystyle\int_2^3 3^x\,dx$.

Solution:

1. Table form: $\int a^x dx = \frac{a^x}{\log a}$, so $F(x) = \frac{3^x}{\log 3}$.
2. Evaluate: $\frac{3^3 - 3^2}{\log 3} = \frac{27 - 9}{\log 3}$.

Answer: $\dfrac{18}{\log 3}$.

Q3. Find $\displaystyle\int \frac{\sin^2 x - \cos^2 x}{\sin x\cos x}\,dx$.

Solution:

1. Split the fraction: $\frac{\sin^2x}{\sin x\cos x} - \frac{\cos^2x}{\sin x\cos x} = \tan x - \cot x$.
2. Integrate: $\log|\sec x| - \log|\sin x|$.

Answer: $\log\left|\sec x\right| - \log\left|\sin x\right| + C$.

Q4. Find $\displaystyle\int \frac{dx}{x^2 + 16}$.

Solution:

1. Match the standard form $\int\frac{dx}{x^2 + a^2} = \frac1a\tan^{-1}\frac xa$ with $a = 4$.

Answer: $\dfrac{1}{4}\tan^{-1}\dfrac{x}{4} + C$.

Q5. Evaluate $\displaystyle\int_0^{\pi/4} \left(\sin x + \cos x\right)dx$.

Solution:

1. Anti-derivative: $-\cos x + \sin x$.
2. Evaluate: $\left(-\frac{1}{\sqrt2} + \frac{1}{\sqrt2}\right) - (-1 + 0) = 0 + 1$.

Answer: $1$.

3-Mark Questions

Q6. Find $\displaystyle\int x\sin 3x\,dx$.

Solution:

1. By parts ($f = x$, $g = \sin 3x$): $\int x\sin 3x\,dx = x\cdot\left(-\frac{\cos 3x}{3}\right) + \frac13\int\cos 3x\,dx$.
2. Finish: $-\frac{x\cos 3x}{3} + \frac{\sin 3x}{9}$.

Answer: $-\dfrac{x\cos 3x}{3} + \dfrac{\sin 3x}{9} + C$.

Q7. Find $\displaystyle\int \frac{dx}{(x + 1)(x + 2)^2}$.

Solution:

1. Repeated-factor template: $\frac{1}{(x+1)(x+2)^2} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{(x+2)^2}$.
2. Find constants: $x = -1$: $A = 1$. $x = -2$: $C = -1$. Compare $x^2$ coefficients: $A + B = 0$, so $B = -1$.
3. Integrate: $\log|x+1| - \log|x+2| + \frac{1}{x+2}$.

Answer: $\log\left|\dfrac{x + 1}{x + 2}\right| + \dfrac{1}{x + 2} + C$.

Q8. Find $\displaystyle\int \sin^{-1}x\,dx$.

Solution:

1. Constant-function trick ($f = \sin^{-1}x$, $g = 1$): $x\sin^{-1}x - \int\frac{x}{\sqrt{1 - x^2}}dx$.
2. Substitute $t = 1 - x^2$ in the remainder: $\int\frac{x\,dx}{\sqrt{1-x^2}} = -\sqrt{1 - x^2}$.

Answer: $x\sin^{-1}x + \sqrt{1 - x^2} + C$.

Q9. Find $\displaystyle\int \frac{3x + 5}{x^2 + 4x + 13}\,dx$.

Solution:

1. Split the numerator along the derivative of the denominator ($2x + 4$): $3x + 5 = \frac32(2x + 4) - 1$.
2. First piece: $\frac32\int\frac{2x+4}{x^2+4x+13}dx = \frac32\log\left(x^2 + 4x + 13\right)$.
3. Second piece: complete the square: $x^2 + 4x + 13 = (x+2)^2 + 9$, so $-\int\frac{dx}{(x+2)^2 + 9} = -\frac13\tan^{-1}\frac{x+2}{3}$.

Answer: $\dfrac{3}{2}\log\left(x^2 + 4x + 13\right) - \dfrac{1}{3}\tan^{-1}\left(\dfrac{x + 2}{3}\right) + C$.

Q10. Evaluate $\displaystyle\int_0^{\pi/2} x\cos x\,dx$.

Solution:

1. By parts: $\int x\cos x\,dx = x\sin x + \cos x$.
2. Evaluate: $\left[x\sin x + \cos x\right]_0^{\pi/2} = \left(\frac{\pi}{2} + 0\right) - (0 + 1)$.

Answer: $\dfrac{\pi}{2} - 1$.

Q11. Evaluate $\displaystyle\int_0^4 |x - 2|\,dx$.

Solution:

1. Split at the kink $x = 2$ (property P2): $\int_0^2(2 - x)dx + \int_2^4(x - 2)dx$.
2. Evaluate each: both are triangles of base $2$ and height $2$: $2 + 2$.

Answer: $4$.

5-Mark Questions

Q12. Evaluate $\displaystyle\int_0^{\pi} \frac{x\tan x}{\sec x + \tan x}\,dx$.

Solution:

1. King's rule (P4): $\tan(\pi - x) = -\tan x$ and $\sec(\pi - x) = -\sec x$, so $\frac{\tan(\pi - x)}{\sec(\pi - x) + \tan(\pi - x)} = \frac{\tan x}{\sec x + \tan x}$ — the trig block is invariant. Hence $I = \int_0^{\pi}\frac{(\pi - x)\tan x}{\sec x + \tan x}dx$.
2. Add: $2I = \pi\int_0^{\pi}\frac{\tan x}{\sec x + \tan x}dx = \pi\int_0^{\pi}\frac{\sin x}{1 + \sin x}dx$.
3. Rewrite: $\frac{\sin x}{1 + \sin x} = 1 - \frac{1}{1 + \sin x}$; and $\int\frac{dx}{1 + \sin x} = \int\frac{1 - \sin x}{\cos^2x}dx = \tan x - \sec x$.
4. Evaluate on $[0, \pi]$: $\int_0^\pi 1\,dx = \pi$; $\left[\tan x - \sec x\right]_0^{\pi} = (0 + 1) - (0 - 1) = 2$.
5. Assemble: $2I = \pi(\pi - 2)$, so $I = \frac{\pi}{2}(\pi - 2)$.

Answer: $\dfrac{\pi^2}{2} - \pi$.

Q13. Find $\displaystyle\int \frac{x^2 + x + 1}{(x + 2)\left(x^2 + 1\right)}\,dx$.

Solution:

1. Template with a quadratic factor: $\frac{x^2+x+1}{(x+2)(x^2+1)} = \frac{A}{x+2} + \frac{Bx + C}{x^2+1}$.
2. Find $A$: put $x = -2$: $\frac{4 - 2 + 1}{4 + 1} = \frac35 = A$.
3. Find $B, C$: comparing $x^2$: $A + B = 1 \Rightarrow B = \frac25$; constants: $A + 2C = 1 \Rightarrow C = \frac15$.
4. Integrate: $\frac35\log|x+2| + \frac25\cdot\frac12\log(x^2+1) + \frac15\tan^{-1}x$.

Answer: $\dfrac{3}{5}\log|x + 2| + \dfrac{1}{5}\log\left(x^2 + 1\right) + \dfrac{1}{5}\tan^{-1}x + C$.

Q14. Evaluate $\displaystyle\int_0^1 \frac{\log(1 + x)}{1 + x^2}\,dx$.

Solution:

1. Substitute $x = \tan\theta$: $dx = \sec^2\theta\,d\theta$, limits $0 \to 0$, $1 \to \frac{\pi}{4}$: $I = \int_0^{\pi/4}\log(1 + \tan\theta)\,d\theta$.
2. King's rule (P4): $1 + \tan\left(\frac{\pi}{4} - \theta\right) = \frac{2}{1 + \tan\theta}$.
3. Add the two copies: $2I = \int_0^{\pi/4}\log 2\,d\theta = \frac{\pi}{4}\log 2$.

Answer: $\dfrac{\pi}{8}\log 2$.

Q15. Find $\displaystyle\int \frac{dx}{\sin x + \sin 2x}$.

Solution:

1. Factor the denominator: $\sin x + \sin 2x = \sin x(1 + 2\cos x)$.
2. Multiply and divide by $\sin x$: $I = \int\frac{\sin x\,dx}{\sin^2x(1 + 2\cos x)} = \int\frac{\sin x\,dx}{(1 - \cos^2x)(1 + 2\cos x)}$.
3. Substitute $t = \cos x$: $I = -\int\frac{dt}{(1 - t)(1 + t)(1 + 2t)}$.
4. Decompose by cover-up: $\frac{1}{(1-t)(1+t)(1+2t)} = \frac{A}{1-t} + \frac{B}{1+t} + \frac{C}{1+2t}$ with $A = \frac16$ (at $t=1$), $B = -\frac12$ (at $t=-1$), $C = \frac43$ (at $t=-\frac12$).
5. Integrate and return $t = \cos x$ (remember $\int\frac{dt}{1-t} = -\log|1-t|$ and the overall minus from step 3): the three pieces assemble to $\frac16\log|1 - \cos x| + \frac12\log|1 + \cos x| - \frac23\log|1 + 2\cos x|$.

Answer: $\dfrac{1}{6}\log\left|1 - \cos x\right| + \dfrac{1}{2}\log\left|1 + \cos x\right| - \dfrac{2}{3}\log\left|1 + 2\cos x\right| + C$.

Q16. Evaluate $\displaystyle\int_0^{\pi/2} \left(2\log\sin x - \log\sin 2x\right)dx$.

Solution:

1. Expand $\log\sin 2x$: $\log\sin 2x = \log 2 + \log\sin x + \log\cos x$.
2. Simplify the integrand: $2\log\sin x - \log\sin 2x = \log\sin x - \log\cos x - \log 2$.
3. Kill the first two by symmetry: by P4, $\int_0^{\pi/2}\log\sin x\,dx = \int_0^{\pi/2}\log\cos x\,dx$, so their difference integrates to $0$.
4. What remains: $-\int_0^{\pi/2}\log 2\,dx = -\frac{\pi}{2}\log 2$.

Answer: $-\dfrac{\pi}{2}\log 2$.