NCERT Class 12 Mathematics — Integrals, Exercise 7.6. All 24 questions solved.

Exercise 7.6 is integration by parts, the technique for products where no substitution helps:

$$\int u\,v\,dx = u\int v\,dx – \int \left[\frac{du}{dx}\int v\,dx\right]dx$$

Everything depends on choosing which factor is $u$. The usual guide is ILATE — Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential — take whichever comes first in that list as $u$.

Key insight. ILATE works because it puts the function that gets simpler when differentiated into the $u$ slot. $\log x$ becomes $\tfrac1x$; $\tan^{-1}x$ becomes $\tfrac{1}{1+x^2}$; $x^n$ drops a power. Choose the other way round and the new integral is worse than the one you started with.

There is also a shortcut worth knowing on sight:

$$\int e^x\big[f(x) + f'(x)\big]dx = e^x f(x) + \mathrm{C}$$

Questions 16 to 20 are all this pattern, and spotting it turns each into a single line.

Questions 1 to 22 — integrate the functions.

Question 1

$x\sin x$

Solution. Algebraic before Trigonometric, so $u = x$:

$$= -x\cos x + \int \cos x\,dx$$

Question 2

$x\sin 3x$

Solution. Same choice, with the chain-rule factor of $3$:

$$= -\frac{x\cos 3x}{3} + \frac{1}{3}\int \cos 3x\,dx$$

Question 3

$x^2 e^x$

Solution. $u = x^2$, and parts must be applied twice — once to reduce $x^2$ to $2x$, again to reduce $2x$ to $2$:

$$= x^2 e^x – 2\int x e^x dx = x^2 e^x – 2\left(x e^x – e^x\right)$$

Question 4

$x\log x$

Solution. Logarithmic beats Algebraic, so $u = \log x$ — even though $\log x$ looks like the harder factor:

$$= \frac{x^2}{2}\log x – \int \frac{x^2}{2}\cdot\frac{1}{x}dx = \frac{x^2}{2}\log x – \frac{x^2}{4}$$

Question 5

$x\log 2x$

Solution. Identical to question 4, since $\tfrac{d}{dx}\log 2x = \tfrac1x$ as well — the constant inside the logarithm disappears on differentiating.

Question 6

$x^2\log x$

Solution. $u = \log x$ again:

$$= \frac{x^3}{3}\log x – \int \frac{x^3}{3}\cdot\frac{1}{x}dx = \frac{x^3}{3}\log x – \frac{x^3}{9}$$

Question 7

$x\sin^{-1}x$

Solution. Inverse trigonometric comes first in ILATE, so $u = \sin^{-1}x$:

$$= \frac{x^2}{2}\sin^{-1}x – \frac{1}{2}\int \frac{x^2}{\sqrt{1-x^2}}dx$$

The remaining integral is handled by writing $x^2 = -(1-x^2) + 1$, which splits it into a square root and a standard arcsine.

Question 8

$x\tan^{-1}x$

Solution. $u = \tan^{-1}x$:

$$= \frac{x^2}{2}\tan^{-1}x – \frac{1}{2}\int \frac{x^2}{1+x^2}dx$$

and $\tfrac{x^2}{1+x^2} = 1 – \tfrac{1}{1+x^2}$.

Question 9

$x\cos^{-1}x$

Solution. The mirror of question 7. Since $\tfrac{d}{dx}\cos^{-1}x = -\tfrac{1}{\sqrt{1-x^2}}$, every sign flips.

Question 10

$\left(\sin^{-1}x\right)^2$

Solution. There is no obvious second factor, so take $1$ as it — the standard trick for a lone inverse function:

$$= x\left(\sin^{-1}x\right)^2 – \int \frac{2x\sin^{-1}x}{\sqrt{1-x^2}}dx$$

The remaining integral needs parts once more, with $u = \sin^{-1}x$.

Question 11

$\dfrac{x\cos^{-1}x}{\sqrt{1-x^2}}$

Solution. Take $u = \cos^{-1}x$ and note that $\tfrac{x}{\sqrt{1-x^2}}$ integrates to $-\sqrt{1-x^2}$:

$$= -\sqrt{1-x^2}\,\cos^{-1}x – \int \sqrt{1-x^2}\cdot\frac{1}{\sqrt{1-x^2}}dx$$

The two square roots cancel in the remaining integral, leaving $\int 1\,dx = x$, which is subtracted:

$$= -\sqrt{1-x^2}\,\cos^{-1}x – x$$

Checking by differentiation:

$$\frac{d}{dx}\left(-\sqrt{1-x^2}\cos^{-1}x – x\right) = \frac{x\cos^{-1}x}{\sqrt{1-x^2}} + 1 – 1 = \frac{x\cos^{-1}x}{\sqrt{1-x^2}} \ \checkmark$$

A note on the printed answer. The answer at the back of the textbook gives $-\sqrt{1-x^2}\cos^{-1}x \;\mathbf{+}\; x + \mathrm{C}$, with a plus sign. That cannot be right: differentiating it gives

$$\frac{x\cos^{-1}x}{\sqrt{1-x^2}} + 2$$

which exceeds the integrand by $2$ at every value of $x$. The sign on the $x$ term is a misprint; the answer above differentiates back exactly.

Question 12

$x\sec^2 x$

Solution. $u = x$, and $\int\sec^2 x\,dx = \tan x$:

$$= x\tan x – \int \tan x\,dx = x\tan x + \log|\cos x|$$

Question 13

$\tan^{-1}x$

Solution. Take $1$ as the second function:

$$= x\tan^{-1}x – \int \frac{x}{1+x^2}dx$$

Question 14

$x(\log x)^2$

Solution. $u = (\log x)^2$, and parts is needed twice:

$$= \frac{x^2}{2}(\log x)^2 – \int x\log x\,dx$$

The remaining integral is question 4.

Question 15

$\left(x^2 + 1\right)\log x$

Solution. $u = \log x$, and $\int\left(x^2+1\right)dx = \tfrac{x^3}{3} + x$:

$$= \left(\frac{x^3}{3} + x\right)\log x – \int \left(\frac{x^2}{3} + 1\right)dx$$

Question 16

$e^x(\sin x + \cos x)$

Solution. This is the $e^x\big[f + f’\big]$ pattern with $f(x) = \sin x$, since $f'(x) = \cos x$:

Question 17

$\dfrac{x e^x}{(1+x)^2}$

Solution. Split the fraction to expose the pattern. Writing $x = (1+x) – 1$:

$$\frac{x}{(1+x)^2} = \frac{1}{1+x} – \frac{1}{(1+x)^2}$$

Now $f(x) = \tfrac{1}{1+x}$ has $f'(x) = -\tfrac{1}{(1+x)^2}$, so the integrand is exactly $e^x\big[f + f’\big]$.

Question 18

$e^x\left(\dfrac{1 + \sin x}{1 + \cos x}\right)$

Solution. Convert to half angles: $1 + \cos x = 2\cos^2\tfrac{x}{2}$ and $1 + \sin x = 1 + 2\sin\tfrac{x}{2}\cos\tfrac{x}{2}$.

$$\frac{1 + \sin x}{1 + \cos x} = \frac{1}{2}\sec^2\frac{x}{2} + \tan\frac{x}{2}$$

With $f(x) = \tan\tfrac{x}{2}$ we get $f'(x) = \tfrac12\sec^2\tfrac{x}{2}$, so this is the pattern again.

Question 19

$e^x\left(\dfrac{1}{x} – \dfrac{1}{x^2}\right)$

Solution. $f(x) = \tfrac1x$ gives $f'(x) = -\tfrac{1}{x^2}$ — the pattern is already written out for you.

Question 20

$\dfrac{(x-3)e^x}{(x-1)^3}$

Solution. Write the numerator in terms of $x – 1$: $x – 3 = (x-1) – 2$.

$$\frac{x-3}{(x-1)^3} = \frac{1}{(x-1)^2} – \frac{2}{(x-1)^3}$$

With $f(x) = \tfrac{1}{(x-1)^2}$ we have $f'(x) = -\tfrac{2}{(x-1)^3}$.

Question 21

$e^{2x}\sin x$

Solution. Neither factor simplifies on differentiating, so parts applied twice returns the original integral. Call it $\mathrm I$ and solve for it:

$$\mathrm I = \frac{e^{2x}\sin x}{2} – \frac{1}{2}\int e^{2x}\cos x\,dx$$

Applying parts to the second integral brings $\mathrm I$ back with a coefficient, and rearranging gives

$$\mathrm I = \frac{e^{2x}}{5}\left(2\sin x – \cos x\right)$$

The “solve for I” move is what makes this class of integral work at all.

Question 22

$\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)$

Solution. Simplify before integrating. Putting $x = \tan\theta$, the standard identity gives

$$\sin^{-1}\left(\frac{2\tan\theta}{1+\tan^2\theta}\right) = \sin^{-1}(\sin 2\theta) = 2\theta = 2\tan^{-1}x$$

So the integrand is just $2\tan^{-1}x$, and question 13 supplies the rest:

$$2\left[x\tan^{-1}x – \frac{1}{2}\log\left(1+x^2\right)\right]$$

Questions 23 and 24 — choose the correct answer.

Question 23

$\displaystyle\int x^2 e^{x^3}dx$ equals

Solution. Despite sitting in the integration-by-parts exercise, this is a plain substitution — $x^2$ is a multiple of the derivative of $x^3$. With $t = x^3$, $dt = 3x^2 dx$:

$$\frac{1}{3}\int e^t dt = \frac{1}{3}e^{x^3}$$

Question 24

$\displaystyle\int e^x\sec x(1 + \tan x)\,dx$ equals

Solution. Expand the bracket:

$$e^x\sec x(1 + \tan x) = e^x\left[\sec x + \sec x\tan x\right]$$

Now $f(x) = \sec x$ has $f'(x) = \sec x\tan x$, so this is the $e^x\big[f + f’\big]$ pattern once more.

Common mistakes

• Choosing $u$ the wrong way round. In question 4, taking $u = x$ leaves $\int\tfrac{x^2}{2}\cdot\tfrac1x\,dx$ — worse than the original. ILATE puts $\log x$ first for a reason.
• Stopping after one application. Questions 3, 10 and 14 need parts twice. Question 21 needs it twice and then solving for $\mathrm I$.
• Missing the $e^x\big[f + f’\big]$ pattern. Questions 16 to 20 are one line each once you see it, and a page of work if you do not. When the integrand has an $e^x$ multiplying a sum, always check whether one term is the derivative of the other.
• Not simplifying first. Question 22’s integrand is $2\tan^{-1}x$; attempting parts on the arcsine as written is far harder.
• Forgetting the sign on $\cos^{-1}$. Its derivative is $-\tfrac{1}{\sqrt{1-x^2}}$, which flips every term in questions 9 and 11 — and is the source of the misprint in the textbook’s answer to question 11.
• Assuming a question belongs to the current technique. Question 23 sits in the parts exercise but is solved by substitution in one step.

Practise next

• Exercise 7.7 — integrals of $\sqrt{x^2 \pm a^2}$ and $\sqrt{a^2 – x^2}$, derived by parts.
• Exercise 7.5 — worth revisiting, since several answers here reduce to partial-fraction forms.