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Integrals NCERT Solutions for Class 12 Maths Exercise 7.2




In this page we have Integrals NCERT Solutions for Class 12 Maths Exercise 7.2 for EXERCISE 7.2 . Hope you like them and do not forget to like , social share and comment at the end of the page.

Integrate the fxnctions in Exercises 1 to 37

Question 1:
$\frac{2 x}{1 + x^{2}}$
Solution
let, 1 + x 2 = u
Then Differentiating both the sides 2x dx = du
Now $\int \frac{2x}{1 + x^{2}} = \int \frac{1}{u}\; du \\ log \left | u \right | + C \\= log \left | 1 + x^{2} \right | + C \\= log (1 + x^{2}) + C$

Question 2:
$\frac{(log\; x) ^{2}}{x}$
Solution
let, $log \left | x \right | = u$
$log \left | x \right | = u \\ \frac{1}{x} dx = du $
Now $\int \frac{(log \left | x \right |)^{2}}{x} dx = \int u^{2} du \\ = \frac{u ^{3}}{3} + C \\ = \frac{(log \left | x \right |) ^{3}}{3} + C$
Question 3:
$\frac{1}{x + x\; log\; x}$
Solution
$\frac{1}{x + x\; log\; x} = \frac{1}{x (1 + log\; x)}$ let, 1 + log x = u
$\frac{1}{x} dx = du $
$\int \frac{1}{x (1 + log\; x)} dx = \int \frac{1}{u} du = log \left | u \right | + C \\ = log \left | 1 + log x \right | + C$

Question 4:
$sin\; x . sin (cos\; x)$
Solution
let, cos x = u
Then -sin x dx = du
$\int sin\; x . sin (cos\; x) dx = - \int sin\; u du \\ = - [- cos u] + C \\ = cos u + C \\ = cos (cos x) + C$

Question 5:
$sin\; (ax + b) cos\; (ax + b)$
Solution
Now this can be re-written as
$sin\; (ax + b) cos\; (ax + b) = \frac{2 sin\; (ax + b) cos\; (ax + b)}{2} = \frac{sin 2 (ax + b)}{2} $
Let, $2 (ax + b) = z \\ 2 a dx = dz $
\int \frac{sin 2 (ax + b)}{2} dr = \frac{1}{2} \int \frac{sin\; z\; dz}{2a} \\ = \frac{1}{4a} [- cos z] + C \\ = - \frac{1}{4a} cos 2 (ax + b) + C$

Question 6:
$\sqrt{ax + b}$
Solution
let, ax + b = z
a dx = dz
or dx = (1/a) dz $\int (ax + b)^{\frac{1}{2}} dx = \frac{1}{a} \int z ^{\frac{1}{2}} dz $ $= \frac{1}{a} (\frac{z ^{\frac{3}{2}}}{\frac{3}{2}}) + C \\ =\frac{2}{3a} (ax + b)^{\frac{3}{2}} + C$

Question 7:
$x \sqrt{x + 2}$
Solution
let, x + 2 = t
then dx = dt
$\int x \sqrt{x + 2} dx = \int (t - 2) \sqrt{t} dt $ $ = \int (t ^{\frac{3}{2}} - 2t ^{\frac{1}{2}}) dt \\ = \int t ^{\frac{3}{2}} dt - 2 \int t ^{\frac{1}{2}}) dt $ $ = \frac{t ^{\frac{5}{2}}}{\frac{5}{2}} - 2 \frac{t ^{\frac{3}{2}}}{\frac{3}{2}} + C \\ = \frac{2}{5} t ^{\frac{5}{2}} - \frac{4}{3} t ^{\frac{3}{2}} + C $ $ = \frac{2}{5} (x + 2) ^{\frac{5}{2}} - \frac{4}{3} (x + 2) ^{\frac{3}{2}} + C \\$

Question 8:
$x \sqrt{1 + 2 x ^{2}}$
Solution
let, 1 + 2 x2 = t
4x dx = dt
$\int x \sqrt{1 + 2 x ^{2}} dx = \int \frac{\sqrt{t}}{4} dt$ $ = \frac{1}{4} \int t ^{\frac{1}{2}} dt \\ = \frac{1}{4} (\frac{t ^{\frac{3}{2}}}{\frac{3}{2}}) + C \\ = \frac{1}{6} (1 + 2 x^{2}) ^{\frac{3}{2}} + C$

Question 9:
$(4x + 2) \sqrt{x ^{2} + x + 1}$
Solution
let, x 2 + x + 1 = t
(2x + 1) dx = dt
$\int (4 x + 2) \sqrt{x ^{2} + x + 1} dx = \int 2 \sqrt{t} dt $ $ = 2 (\frac{t ^{\frac{3}{2}}}{\frac{3}{2}}) + C \\ = \frac{4}{3} (x ^{2} + x + 1) ^{\frac{3}{2}} + C$

Question 10:
$\frac{1}{x - \sqrt{x}}$
Solution
It can be re-written as $\frac{1}{x – \sqrt{x}} = \frac{1}{\sqrt{x} (\sqrt{x} – 1)}$
let,
$\sqrt{x} - 1 = t$ $ \frac{1}{2 \sqrt{x}} dx = dt$ $\int \frac{1}{\sqrt{x} (\sqrt{x} – 1)} dx = \int \frac{2}{t} dt \\= 2 log \left | z \right | + C \\ =2 log \left | \sqrt{x} – 1 \right | + C$

Question 11:
$\frac{x}{\sqrt{x + 4}},$ x > 0
Solution
let, x + 4 = t
dx = dt
$\int \frac{x}{\sqrt{x + 4}} dx = \int \frac{(t – 4)}{\sqrt{t}} dt $ $ = \int (\sqrt{t} – \frac{4}{\sqrt{t}}) dt \\ = \frac{t ^{\frac{3}{2}}}{\frac{3}{2}} – 4 (\frac{t ^{\frac{1}{2}}}{\frac{1}{2}}) + C \\ = \frac{2}{3} t ^{\frac{3}{2}} – 8 t ^{\frac{1}{2}} + C \\ = \frac{2}{3} t. t ^{\frac{1}{2}} – 8 t ^{\frac{1}{2}} + C \\ = \frac{2}{3} t ^{\frac{1}{2}} (t – 12) + C \\ = \frac{2}{3} (x + 4) ^{\frac{1}{2}} (x + 4 – 12) + C \\ = \frac{2}{3} \sqrt{(x + 4)} (x – 8) + C$

Question 12:
$(x ^{3} – 1) ^{\frac{1}{3}} x ^{5}$
Solution
let, x 3 - 1 = t
3 x 2 = dt
$\int (x ^{3} – 1) ^{\frac{1}{3}} x ^{5} dx = \int (x ^{3} – 1) ^{\frac{1}{3}} x ^{3} . x ^{2} dx \\ = \int t ^{\frac{1}{3}} (t + 1) \frac{dt}{3} \\ = \frac{1}{3} \int (t ^{\frac{4}{3}} + t ^{\frac{1}{3}}) dt \\ = \frac{1}{3} [\frac{t ^{\frac{7}{3}}}{\frac{7}{3}} + \frac{t ^{\frac{4}{3}}}{\frac{4}{3}}] + C \\ = \frac{1}{3} [\frac{3}{7} t ^{\frac{7}{3}} + \frac{3}{4} t ^{\frac{4}{3}}] + C \\ = \frac{1}{7} (x ^{3} – 1) ^{\frac{7}{3}} + \frac{1}{4} (x ^{3} – 1) ^{\frac{4}{3}}] + C$

Question 13:
$\frac{x ^{2}}{(2 + 3x ^{3}) ^{3}} \\$
Solution
let, $2 + 3x ^{3} = z \\ 9 x ^{2} dx = dz $
$ \int \frac{x ^{2}}{(2 + 3 x ^{3})} dx = \frac{1}{9} \int \frac{dz}{(z) ^{3}} \\ = \frac{1}{9} \int {(z) ^{- 3}} dz \\ = \frac{1}{9} (\frac{z ^{- 2}}{- 2}) + C \\ = – \frac{1}{18} (\frac{1}{z ^{2}}) + C \\ = \frac{- 1}{18 (2 + 3x ^{3}) ^{2}} + C \\$


Question 14:
$\frac{1}{x (log x) ^{m}}, x > 0 \\$
Solution
let, $log x = z \\ \frac{1}{x} dx = dz $ $ \int \frac{1}{x (log x) ^{m}} dx = \int \frac{dz}{z ^{m}} \\ = \int z ^{- m} dz \\ = \frac{z ^{- m + 1}}{- m + 1} + C \\ = \frac{z ^{1 – m}}{1 -m} + C \\ = \frac{log x ^{1 – m}}{1 -m} + C$


Question 15:
$\frac{x}{9 – 4 x ^{2}}$
Solution
let, $9 – 4 x ^{2} = t \\ – 8 x dx = dt \\ \int \frac{x}{9 – 4 x ^{2}} = – \frac{1}{8} \int \frac{1}{t} dt \\ = – \frac{1}{8} log \left | t \right | + C \\ = – \frac{1}{8} log \left | 9 – 4 x ^{2} \right | + C$


Question 16:
$e ^{2 x + 3}$
Solution
let, ${2 x + 3} = t \\ 2 dx = dt $
$ \int e ^{2 x + 3} dx = \frac{1}{2} \int e ^{t} dt \\ = \frac{1}{2} (e ^{t}) + K \\ = \frac{1}{2} (e ^{2 x + 3}) + K$

Question 17:
$\frac{x}{e ^{x ^{2}}}$
Solution
let, x 2 = t
2x dx = dt
$\int \frac{x}{e ^{x ^{2}}} dx = \frac{1}{2} \int \frac{1}{e ^{t}} dt \\ = \frac{1}{2} \int e ^{- t} dt \\ = \frac{1}{2} \frac{e ^{- t}}{- 1} + C \\ = – \frac{1}{2} e ^{- x ^{2}} + C \\ = – \frac{1}{2 e ^{x ^{2}}} + C$
Question 18:
$\frac{e ^{tan ^{- 1} x}}{1 + x ^{2}}$
Solution
let, $tan ^{- 1} x = z$
$ \frac{1}{1 + x ^{2}} dx = dz $
$ \int \frac{e ^{tan ^{- 1} x}}{1 + x ^{2}} dx = \int e ^{z} dz \\ = e ^{z} + C \\ = e ^{tan ^{- 1} x} + C$

Question 19:
$\frac{e ^{2x} – 1}{e ^{2x} + 1}$
Solution
$\frac{e ^{2x} – 1}{e ^{2x} + 1}$
Dividing the nxmerator and denominator by e x, we get
$\frac{\frac{e ^{2x} – 1}{e ^{x}}}{\frac{e ^{2x} + 1}{e ^{x}}} = \frac{e ^{x} – e ^{- x}}{e ^{x} + e ^{- x}} \\$
let,
$e ^{x} + e ^{- x} = z \\ (e ^{x} – e ^{- x}) dx = dz $
$ \int \frac{e ^{2x} – 1}{e ^{2x} + 1} dx = \int \frac{e ^{x} – e ^{- x}}{e ^{x} + e ^{- x}} dx \\ = \int \frac{dz}{z} \\ = log \left | z \right | + C \\ = log \left | e ^{x} + e ^{- x} \right | + C$

Question 20:
$\frac{e ^{2x} – e ^{- 2x}}{e ^{2x} + e ^{- 2x}}$
Solution
let, $e ^{2x} + e ^{- 2x} = z \\ (2 e ^{2x} – 2 e ^{- 2x}) dx = dz \\ 2 (e ^{2x} – e ^{- 2x}) dx = dz \\ \int \frac{e ^{2x} – e ^{- 2x}}{e ^{2x} + e ^{- 2x}} = \int \frac{dz}{2z} dz \\ = \frac{1}{2} \int \frac{1}{z} dz \\ = \frac{1}{2} log \left | z \right | + C \\ = \frac{1}{2} log \left | e ^{2x} + e ^{- 2x} \right | + C$


Question 21:
$tan ^{2} (2 x – 3)$
Solution
we know that, $tan ^{2} (2 x – 3) = sec ^{2} (2 x – 3) – 1$
let, $2 x – 3 = t \\ 2 dx = dt$
$ \int tan ^{2} (2 x – 3) dx = \int [sec ^{2} (2 x – 3) – 1] dx \\ = \frac{1}{2} \int (sec ^{2} t) dt – \int 1 d t \\ = \frac{1}{2} tan t – x + C \\ = \frac{1}{2} tan (2 x – 3) – x + C$

Question 22:
$sec ^{2} (7 – 4 x)$
Solution
let, $(7 – 4 x) = t \\ – 4\; d x = dt$ $\int sec ^{2} (7 – 4 x) dx = – \frac{1}{4} \int sec ^{2} t dt \\ = – \frac{1}{4} (tan t) + C \\ = – \frac{1}{4} [tan (7 – 4 x)] + C$

Question 23:
$\frac{sin ^{- 1} x}{\sqrt{1 – x ^{2}}}$
Solution
let, $sin ^{- 1} x = z \\ \frac{1}{\sqrt{1 – x ^{2}}} dx = dz \\ \int \frac{sin ^{- 1} x}{\sqrt{1 – x ^{2}}} dx = \int z dz \\ = \frac{z ^{2}}{2} + C \\ = \frac{(sin ^{- 1} x) ^{2}}{2} + C \\$

Question 24:
$\frac{2 cos\; x – 3 sin\; x}{6 cos\; x + 4 sin\; x}$
Solution
$\frac{2 cos\; x – 3 sin\; x}{6 cos\; x + 4 sin\; x} = \frac{2 cos\; x – 3 sin\; x}{2 (3 cos\; x + 2 sin\; x)}$ let,
$3\; cos\; x + 2\; sin\; x = z\\ (- 3\; sin\; x + 2\; cos\; x) dx = dt $
$\int \frac{2 cos\; x – 3 sin\; x}{6 cos\; x + 4 sin\; x} dx = \int \frac{dt}{2t} \\ = \frac{1}{2} \frac{1}{t} dt \\ = \frac{1}{2} log \left | t \right | + C \\ = \frac{1}{2} log \left | 3\; cos\; x + 2\; sin\; x \right | + C \\$


Question 25:
$\frac{1}{cos ^{2} x (1 – tan x) ^{2}}$
Solution
$\frac{1}{cos ^{2} x (1 – tan x) ^{2}} = \frac{sec ^{2} x}{(1 – tan x) ^{2}} \\$ let,
$(1 – tan x) = t \\ sec ^{2} x dx = dt $
$\int \frac{sec ^{2} x}{(1 – tan x) ^{2}} dx = \int -\frac{dt}{t ^{2}} \\ = – \int t ^{- 2} dt \\ = \frac{1}{t} + C \\ = \frac{1}{1 – tan x} + C$

Question 26:
$\frac{cos \sqrt{x }}{\sqrt{x }}$
Solution
let, $\sqrt{x } = t \\ \frac{1}{2 \sqrt{x}} dx = dt $
$ \int \frac{cos \sqrt{x }}{\sqrt{x }} = 2 \int cos\; t dt \\ = 2 sin\; t + C \\ = 2 sin\; \sqrt{x } + C$


Question 27:
$\sqrt{sin\; 2 x}\; cos\; 2 x$
Solution
let, $sin\; 2 x = t \\ 2 cos\; 2 x dx = dt$
$\int \sqrt{sin\; 2 x}\; cos\; 2 x = \frac{1}{2} \int \sqrt{t} dt \\ = \frac{1}{2} (\frac{t ^{\frac{3}{2}}}{\frac{3}{2}}) + C \\ = \frac{1}{3} t ^{\frac{3}{2}} + C \\ = \frac{1}{3} (sin\; 2 x) ^{\frac{3}{2}} + C$

Question 28:
$\frac{cos\; x}{\sqrt{1 + sin\; x}}$
Solution
let, $1 + sin\; x = t \\ cos\; x dx = dt $
$\int \frac{cos\; x}{\sqrt{1 + sin\; x}} dx = \int \frac{dt}{\sqrt{t}} \\ = 2 \sqrt{t} + C \\ = 2 \sqrt{1 + sin\; x} + C$


Question 29:
$cot\; x\; log\; sin\; x$
Solution
let, $log\; sin\; x = t \\ \frac{1}{sin\; x }. cos\; x = dt \\ cot\; x\; dx = dt $
$ \int cot\; x\; log\; sin\; x dx = \int t\; dt \\ = \frac{t ^{2}}{2} + C \\ = \frac{1}{2} (log\; sin\; x) ^{2} + C$

Question 30:
$\frac{sin\; x}{1 + cos\; x}$
Solution
let,
$1 + cos\; x = z \\ – sin\; x dx = dt $
$\int \frac{sin\; x}{1 + cos\; x} dx = \int – \frac{dt}{t } \\ = – \int \frac{dt}{t} \\ = – log \left | t \right | + C \\ = – log \left | 1 + cos\; x \right | + C$

Question 31:
$\frac{sin\; x}{(1 + cos\; x) ^{2}}$
Solution
let, $1 + cos\; x = t $
$– sin\; x dx = dt $
$\int \frac{sin\; x}{1 + cos\; x} dx = \int – \frac{dt}{t ^{2}} \\ = – \int \frac{dt}{t ^{2}} \\ = \frac{1}{t} + C \\ = \frac{1}{1 + cos\; x} + C$

Question 32:
$\frac{1}{1 + cot\; x}$
Solution
let, I = $\int \frac{1}{1 + cot\; x} dx \\ = \int \frac{1}{1 + \frac{cos\; x}{sin\; x}} dx \\ = \int \frac{sin\; x}{sin\; x + cos\; x} dx \\ = \frac{1}{2} \int \frac{2 sin\; x}{sin\; x + cos\; x} dx \\ = \frac{1}{2} \int \frac{(sin\; x + cos\; x) + (sin\; x – cos\; x)}{(sin\; x + cos\; x)} dx \\ = \frac{1}{2} \int 1 dx + \frac{1}{2} \int \frac{(sin\; x – cos\; x)}{(sin\; x + cos\; x)} dx \\ = \frac{1}{2} x + \frac{1}{2} \int \frac{(sin\; x – cos\; x)}{(sin\; x + cos\; x)} dx \\$
$let,\; (sin\; x + cos\; x) = y \\ = (cos\; x – sin\; x) dx = dy \\ I = \frac{x }{2} + \frac{1}{2} log \left | y \right | + K \\ = \frac{x }{2} – \frac{1}{2} log \left | (sin\; x + cos\; x) \right | + K \\$

Question 33:
$\frac{1}{1 – tan x}$
Solution
let,
$I =\int \frac{1}{1 – tan\; x} dx \\ = \int \frac{1}{1 – \frac{sin\; x}{cos\; x}} dx \\ = \int \frac{cos\; x}{cos\; x – sin\; x} dx \\ = \frac{1}{2} \int \frac{2 cos\; x}{cos\; x – sin\; x} dx \\ = \frac{1}{2} \int \frac{(cos\; x – sin\; x ) + (cos\; x + sin\; x)}{(cos\; x – sin\; x)} dx \\ = \frac{1}{2} \int 1 dx + \frac{1}{2} \int \frac{(cos\; x + sin\; x)}{(cos\; x – sin\; x)} dx \\ = \frac{1}{2} x + \frac{1}{2} \int \frac{(cos\; x + sin\; x)}{(cos\; x – sin\; x)} dx \\$
$let,\; (cos\; x – sin\; x) = y \\ = (- sin\; x – cos\; x ) dx = dy$
Then
$I = \frac{x }{2} – \frac{1}{2} log \left | \y \right | + C \\ = \frac{x }{2} – \frac{1}{2} log \left | (cos\; x – sin\; x) \right | + C \\$
Question 34:
$\frac{\sqrt{tan x}}{sin\; x\; cos\; x}$
Solution
let,
$l I = \frac{\sqrt{tan x}}{sin\; x\; cos\; x} dx \\ = \frac{\sqrt{tan x} \times cos\; x}{sin\; x\; cos\; x \times cos\; x} dx \\ = \int \frac{\sqrt{tan x}}{tan\; x\; cos\; ^{2} x} dx \\ = \int \frac{sec\; ^{2} x}{\sqrt{tan\; x}} dx $
$let, tan\; x = y \\ sec\; ^{2} x dx = dy $
Then
$I = \int \frac{dy}{\sqrt{y}} \\ = 2 \sqrt{y} + K \\ = 2 \sqrt{tan\; x} + K$


Question 35:
$\frac{(1 + log x) ^{2}}{x}$
Solution
let,
$ 1 + log x = y \\ \frac{1}{x} dx = dy $
$ \int \frac{(1 + log x) ^{2}}{x} dx = \int y ^{2}\; dy \\ = \frac{y ^{3}}{3} + C \\ = \frac{(1 + log x) ^{3}}{3} + C$


Question 36:
$\frac{(x + 1)(x + log x) ^{2}}{x}$
Solution
We can write the integral in the form
$\frac{(x + 1)(x + log x) ^{2}}{x} = \frac{(x + 1)}{x} (x + log x) ^{2} = (1 + \frac{1}{x}) (x + log x) ^{2} $
let,$\; (x + log x) = t \\ (1 + \frac{1}{x}) dx = dt $
$ \int (1 + \frac{1}{x}) (x + log\; x) ^{2} dx = \int t ^{2} dt \\ = \frac{t ^{3}}{3} + C \\ = \frac{1}{3} (x + log x) ^{3} + C$

Question 37:
$\frac{x ^{3}\; sin\; (tan ^{- 1} x ^{4})}{1 + x ^{8}}$
Solution
This is a example of double substitution
$let,\; x ^{4} = t \\ 4 x ^{3} dx = dt $
$ \int \frac{x ^{3}\; sin\; (tan ^{- 1} x ^{4})}{1 + x ^{8}} dx = \frac{1}{4} \int \frac{sin\; (tan ^{- 1} t)}{1 + t ^{2}} \;dt …. (1) $
$ let,\; tan ^{- 1} t = y \\ \frac{1}{1 + t ^{2}} dt = du $
So we get
$\int \frac{x ^{3}\; sin\; (tan ^{- 1} x ^{4})}{1 + x ^{8}} dx = \frac{1}{4} \int sin\; u\; du \\ = \frac{1}{4} (- cos u) + c \\ = – \frac{1}{4} cos (tan ^{- 1} t) + C \\ = – \frac{1}{4} cos (tan ^{- 1} x ^{4}) + C \\$

Question 38:
Which of the following below is the answer for $\int \frac{10 x ^{9} + 10 ^{x} log_{e} 10}{x ^{10} + 10 ^{x}} dx$ :
(a)$ 10 ^{x} – x ^{10} + C $
(b) $10 ^{x} + x ^{10} + C $
(c) $(10 ^{x} – x ^{10}) ^{- 1} + C $
(d) $log (10 ^{x} + x ^{10}) + C $
Solution
$x ^{10} + 10 ^{x} = z \\ (10 x ^{9} + 10 ^{x} log_{e} 10) dx = dz $
$\int \frac{10 x ^{9} + 10 ^{x} log_{e} 10}{x ^{10} + 10 ^{x}} dx = \int \frac{dz}{z} \\ = log z + C \\ = log (x ^{10} + 10 ^{x}) + C $
Therefore D is the correct answer

Question 39:
Which of the following below is the answer for $\int \frac{dx}{sin ^{2} \;x\; cos ^{2} \;x}$
(a)$ tan\; x + cot\; x + C $
(b) $tan\; x – cot\; x + C $
(c) $tan\; x \; cot\; x + C $
(d) $tan\; x – cot\; 2x + C $
Solution
$I = \int \frac{dx}{sin ^{2} \;x\; cos ^{2} \;x} \\ = \int \frac{1}{sin ^{2} \;x\; cos ^{2} \;x} dx \\ = \int \frac{sin ^{2} \;x\; + cos ^{2} \;x}{sin ^{2} \;x\; cos ^{2} \;x} dx \\ = \int \frac{sin ^{2} \;x}{sin ^{2} \;x\; cos ^{2} \;x} dx + \int \frac{cos ^{2} \;x}{sin ^{2} \;x\; cos ^{2} \;x} dx \\ = \int sec ^{2} \;x dx + \int cosec ^{2} \;x dx \\ = tan\; x – cot\; x + C \\$ Therefore, B is the correct answer
 

 


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