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Maths Solve the Following Question.(2 Marks)

Indefinite Integration · Maharashtra Board STD 12 (MSBSHSE - Semi English Medium). 94 questions.

Questions · Page 2 of 2

Solve the Following Question.(2 Marks)

Question 602 Marks
Integrate the following functions w.r.t. x:

$\frac{1}{x \cdot \log x \cdot \log (\log x)}$

Answer
Let $I=\int \frac{1}{x \cdot \log x \cdot \log (\log x)} d x$

$=\int \frac{1}{\log (\log x)} \cdot \frac{1}{x \cdot \log x} d x$

Put $\log (\log x)=t \quad \therefore \frac{1}{\log x} \cdot \frac{1}{x} d x=d t$

$\therefore \frac{1}{x \cdot \log x} d x=d t$

$\begin{aligned} \therefore I & =\int \frac{1}{t} d t=\log |t|+c \\ & =\log |\log (\log x)|+c .\end{aligned}$

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Question 612 Marks
Integrate the following functions w.r.t. x:

$\frac{x^2}{\sqrt{9-x^6}}$

Answer
Let $I=\int \frac{x^2}{\sqrt{9-x^6}} d x$

Put $x^3=t \quad \therefore 3 x^2 d x=d t \quad \therefore x^2 d x=\frac{1}{3} d t$

$\therefore I=\int \frac{1}{\sqrt{9-t^2}} \cdot \frac{d t}{3}$

$\begin{aligned} & =\frac{1}{3} \int \frac{d t}{\sqrt{3^2-t^2}} \\ = & \frac{1}{3} \sin ^{-1}\left(\frac{t}{3}\right)+c \\ = & \frac{1}{3} \sin ^{-1}\left(\frac{x^3}{3}\right)+c .\end{aligned}$

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Question 622 Marks
Integrate the following functions w.r.t. x:
$(5-3 x)(2-3 x)^{-\frac{1}{2}}$
Answer
Let $I=\int(5-3 x)(2-3 x)^{-\frac{1}{2}} d x$Put $2-3 x=t$
$\therefore-3 d x=d t$
$\therefore d x=\frac{-d t}{3}$
Also, $x=\frac{2-t}{3}$
$ \therefore I  =\int\left[5-3\left(\frac{2-t}{3}\right)\right] t^{-\frac{1}{2}} \cdot\left(\frac{-d t}{3}\right)$
$=-\frac{1}{3} \int(5-2+t) t^{-\frac{1}{2}} d t$
$ =-\frac{1}{3} \int(3+t) t^{-\frac{1}{2}} d t$
$ =-\frac{1}{3} \int\left(3 t^{-\frac{1}{2}}+t^{\frac{1}{2}}\right) d t$
$=-\frac{3}{3} \int t^{-\frac{1}{2}} d t-\frac{1}{3} \int t^{\frac{1}{2}} d t$
$=-\frac{t^{\frac{1}{2}}}{(1 / 2)}-\frac{1}{3} \cdot \frac{t^{\frac{3}{2}}}{(3 / 2)}+c$
$=-2 \sqrt{2-3 x}-\frac{2}{9}(2-3 x)^{\frac{3}{2}}+c .$
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Question 632 Marks
Integrate the following functions w.r.t. x:
$x^5 \sqrt{a^2+x^2}$
Answer
Let $I=\int x^5 \sqrt{a^2+x^2} d x$Put, $a^2+x^2=t$
$\therefore 2 x d x=d t \quad \therefore x d x=\frac{1}{2} d t$
Also, $x^2=t-a^2$
$I  =\int x^2 \cdot x^2 \sqrt{a^2+x^2} x d x$
$=\frac{1}{2} \int\left(t-a^2\right)^2 \sqrt{t} d t$
$=\frac{1}{2} \int\left(t^2-2 a^2 t+a^4\right) \sqrt{t} d t$
$=\frac{1}{2} \int\left(t^{\frac{5}{2}}-2 a^2 t^{\frac{3}{2}}+a^4 t^{\frac{1}{2}}\right) d t$
$=\frac{1}{2} \int t^{\frac{5}{2}} d t-a^2 \int t^{\frac{3}{2}} d t+\frac{a^4}{2} \int t^{\frac{1}{2}} d t$
$=\frac{1}{2} \cdot \frac{t^{\frac{7}{2}}}{\left(\frac{7}{2}\right)}-a^2 \cdot \frac{t^{\frac{5}{2}}}{\left(\frac{5}{2}\right)}+\frac{a^4}{2} \cdot \frac{t^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}+c$
$=\frac{1}{7}\left(a^2+x^2\right)^{\frac{7}{2}}-\frac{2 a^2}{5}\left(a^2+x^2\right)^{\frac{5}{2}}+\frac{a^4}{3}\left(a^2+x^2\right)^{\frac{3}{2}}+c .$
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Question 642 Marks
Integrate the following functions w.r.t. x:
$(2 x+1) \sqrt{x+2}$
Answer
Let $I=\int(2 x+1) \sqrt{x+2} d x$Put $x+2=t \quad \therefore d x=d t$
Also, $x=t-2$
$ \therefore 2 x+1=2(t-2)+1=2 t-3$
$\therefore I=\int(2 t-3) \sqrt{t} d t$
$=\int\left(2 t^{\frac{3}{2}}-3 t^{\frac{1}{2}}\right) d t$
$=2 \int t^{\frac{3}{2}} d t-3 \int t^{\frac{1}{2}} d t$
$=2 \cdot \frac{t^{\frac{5}{2}}}{\left(\frac{5}{2}\right)}-3 \cdot \frac{t^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}+c$
$=\frac{4}{5}(x+2)^{\frac{5}{2}}-2(x+2)^{\frac{3}{2}}+c$.
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Question 652 Marks
Integrate the following functions w.r.t. x:
$\frac{1}{\sqrt{x}+\sqrt{x^3}}$
Answer
Let $I=\int \frac{1}{\sqrt{x}+\sqrt{x^3}} d x$$=\int \frac{1}{x^{\frac{1}{2}}+x^{\frac{3}{2}}} d x$
Put $x=t^2 \quad \therefore d x=2 t d t$
Also $x^{\frac{1}{2}}=\left(t^2\right)^{\frac{1}{2}}=t$ and $x^{\frac{3}{2}}=\left(t^2\right)^{\frac{3}{2}}=t^3$
$\therefore I  =\int \frac{2 t d t}{t+t^3}$
$ =2 \int \frac{t d t}{t\left(1+t^2\right)}$
$=2 \int \frac{1}{1+t^2} d t$
$=2 \tan ^{-1} t+c$
$=2 \tan ^{-1}(\sqrt{x})+c .$
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Question 662 Marks
Integrate the following functions w.r.t. x:
$\frac{(x-1)^2}{\left(x^2+1\right)^2}$
Answer
Let $I=\int \frac{(x-1)^2}{\left(x^2+1\right)^2} d x$$=\int \frac{x^2-2 x+1}{\left(x^2+1\right)^2} d x$
$=\int \frac{\left(x^2+1\right)-2 x}{\left(x^2+1\right)^2} d x$
$=\int\left[\frac{x^2+1}{\left(x^2+1\right)^2}-\frac{2 x}{\left(x^2+1\right)^2}\right] d x$
$=\int \frac{1}{x^2+1} d x-\int \frac{2 x}{\left(x^2+1\right)^2} d x$
$=I_1-I_2$
In $I_2$, Put $x^2+1=t \quad \therefore 2 x d x=d t$
$ \therefore I  =\int \frac{1}{x^2+1} d x-\int t^{-2} d t$
$ =\tan ^{-1} x-\frac{t^{-1}}{(-1)}+c$
$ =\tan ^{-1} x+\frac{1}{x^2+1}+c .$
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Question 672 Marks
Integrate the following functions w.r.t. x:

$\frac{\sqrt{\tan x}}{\sin x \cdot \cos x}$

Answer
Let $I =\int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} d x$

Dividing numerator and denominator by $\cos ^2 x$, we get

$\begin{aligned} I & =\int \frac{\left(\frac{\sqrt{\tan x}}{\cos ^2 x}\right)}{\left(\frac{\sin x}{\cos x}\right)} d x \\ & =\int \frac{\sqrt{\tan x} \cdot \sec ^2 x}{\tan x} d x\end{aligned}$

$=\int \frac{\sec ^2 x}{\sqrt{\tan x}} d x$

Put $\tan x=t \quad \therefore \sec ^2 x d x=d t$

$\begin{aligned} \therefore I & =\int \frac{1}{\sqrt{t}} d t=\int t^{-\frac{1}{2}} d t \\ & =\frac{t^{\frac{1}{2}}}{1 / 2}+c=2 \sqrt{t}+c \\ & =2 \sqrt{\tan x}+c .\end{aligned}$

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Question 682 Marks
Integrate the following functions w.r.t. x:
$e^{3 \log x} \cdot\left(x^4+1\right)^{-1}$
Answer
Let $I=e^{3 \log x}\left(x^4+1\right)^{-1} d x$
$=\int \frac{e^{\log x^3}}{x^4+1} d x$
$=\int \frac{x^3}{x^4+1} d x \quad \ldots\left[\because e^{\log N}=N\right]$
$=\frac{1}{4} \int \frac{4 x^3}{x^4+1} d x$
$=\frac{1}{4} \int \frac{\frac{d}{d x}\left(x^4+1\right)}{x^4+1} d x$
$=\frac{1}{4} \log \left|x^4+1\right|+c . \ldots\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c\right]$
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Question 692 Marks
Integrate the following functions w.r.t. x:
$x^9 \cdot \sec ^2\left(x^{10}\right)$
Answer
Let $I=\int x^9 \cdot \sec ^2\left(x^{10}\right) d x$
Put $x^{10}=t \quad \therefore 10 x^9 d x=d t \quad \therefore x^9 d x=\frac{1}{10} d t$
$\therefore I=\int \sec ^2 t \cdot \frac{d t}{10}$
$=\frac{1}{10} \int \sec ^2 t d t$
$=\frac{1}{10} \tan t+c$
$=\frac{1}{10} \tan \left(x^{10}\right)+c .$
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Question 702 Marks
Integrate the following functions w.r.t. x:

$\frac{1}{4 x+5 x^{-11}}$

Answer
Let $I=\int \frac{1}{4 x+5 x^{-11}} d x$

$=\int \frac{x^{11}}{x^{11}\left(4 x+5 x^{-11}\right)} d x$

$=\int \frac{x^{11}}{4 x^{12}+5} d x$

$=\frac{1}{48} \int \frac{48 x^{11}}{4 x^{12}+5} d x$

$=\frac{1}{48} \int \frac{\frac{d}{d x}\left(4 x^{12}+5\right)}{4 x^{12}+5} d x$

$=\frac{1}{48} \log \left|4 x^{12}+5\right|+c \ldots\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c\right]$

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Question 712 Marks
Integrate the following functions w.r.t. x:
$\frac{e^{2 x}+1}{e^{2 x}-1}$
Answer
Let $I=\int \frac{e^{2 x}+1}{e^{2 x}-1} d x=\int \frac{\left(\frac{e^{2 x}+1}{e^x}\right)}{\left(\frac{e^{2 x}-1}{e^x}\right)} d x$
$=\int\left(\frac{e^x+e^{-x}}{e^x-e^{-x}}\right) d x$
$=\int \frac{\frac{d}{d x}\left(e^x-e^{-x}\right)}{e^x-e^{-x}} d x$
$=\log \left|e^x-e^{-x}\right|+c \quad \cdots\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c\right]$
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Question 722 Marks
Integrate the following functions w.r.t. x:

$\frac{e^x \cdot \log \left(\sin e^x\right)}{\tan \left(e^x\right)}$

Answer
Let $I=\int \frac{e^x \cdot \log \left(\sin e^x\right)}{\tan \left(e^x\right)} d x$

$=\int \log \left(\sin e^x\right) \cdot e^x \cot \left(e^x\right) d x$

Put $\log \left(\sin e^x\right)=t \quad \therefore \frac{1}{\sin \left(e^x\right)} \cdot \cos \left(e^x\right) \cdot e^x d x=d t$

$\therefore e^x \cdot \cot \left(e^x\right) d x=d t$

$\begin{aligned} \therefore I & =\int t d t=\frac{t^2}{2}+c \\ & =\frac{1}{2}\left[\log \left(\sin e^x\right)\right]^2+c .\end{aligned}$

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Question 732 Marks
Integrate the following functions w.r.t. x:

$\frac{\left(x^2+2\right)}{\left(x^2+1\right)} \cdot a^{x+\tan ^{-1} x}$

Answer
Let $I=\int \frac{x^2+2}{\left(x^2+1\right)} \cdot a^{x+\tan ^{-1} x} d x$

$=\int a^{x+\tan ^{-1} x} \cdot\left(\frac{x^2+2}{x^2+1}\right) d x$

Put $x+\tan ^{-1} x=t$

$\begin{array}{c}\therefore\left(1+\frac{1}{1+x^2}\right) d x=d t \\ \therefore\left(\frac{1+x^2+1}{1+x^2}\right) d x=d t \\ \therefore\left(\frac{x^2+2}{x^2+1}\right) d x=d t \\ \therefore I=\int a^t d t=\frac{a^t}{\log a}+c \\ =\frac{a^{x+\tan ^{-1} x}}{\log a}+c .\end{array}$

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Question 742 Marks
Integrate the following functions w.r.t. x:
$\frac{x \cdot \sec ^2\left(x^2\right)}{\sqrt{\tan ^3\left(x^2\right)}}$
Answer
Let $I=\int \frac{x \cdot \sec ^2\left(x^2\right)}{\sqrt{\tan ^3\left(x^2\right)}} d x$
Put $\tan \left(x^2\right)=t \quad \therefore \sec ^2\left(x^2\right) \times 2 x d x=d t$
$\therefore x \cdot \sec ^2\left(x^2\right) d x=\frac{d t}{2}$
$\therefore I=\int \frac{1}{\sqrt{t^3}} \cdot \frac{d t}{2}=\frac{1}{2} \int t^{-\frac{3}{2}} d t$
$=\frac{1}{2} \cdot \frac{t^{-\frac{1}{2}}}{-1 / 2}+c=\frac{-1}{\sqrt{t}}+c=\frac{-1}{\sqrt{\tan \left(x^2\right)}}+c$.
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Question 752 Marks
Integrate the following functions w.r.t. x:
$\frac{1+x}{x-\sin (x+\log x)}$
Answer
$ \text { Let } I=\int \frac{1+x}{x \cdot \sin (x+\log x)} d x$
$=  \int \frac{1}{\sin (x+\log x)} \cdot\left(\frac{1+x}{x}\right) d x$
$=  \int \frac{1}{\sin (x+\log x)} \cdot\left(\frac{1}{x}+1\right) d x$
Put $x+\log x=t \quad \therefore\left(1+\frac{1}{x}\right) d x=d t$
$\therefore I=\int \frac{1}{\sin t} d t=\int \operatorname{cosec} t d t$
$=\log |\operatorname{cosec} t-\cot t|+c$
$=\log |\operatorname{cosec}(x+\log x)-\cot (x+\log x)|+c .$
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Question 842 Marks
Evaluate:
If $f^{\prime}(x)=x-\frac{3}{x^3}, f(1)=\frac{11}{2}$, find $f(x)$.
Answer
By the definition of integral,
$f(x)=\int f^{\prime}(x) d x$
$=\int\left(x-\frac{3}{x^3}\right) d x$
$=\int x d x-3 \int x^{-3} d x$
$=\frac{x^2}{2}-\frac{3 x^{(-2)}}{(-2)}+c$
$=\frac{x^2}{2}+\frac{3}{2 x^2}+c$
........(1)
$f(1)=\frac{11}{2}$ ....... (Given)
$\therefore \frac{1}{2}+\frac{3}{2}+c=\frac{11}{2}$
$\therefore c=\frac{7}{2}$
$\therefore f(x)=\frac{x^2}{2}+\frac{3}{2 x^2}+\frac{7}{2} . \quad \ldots \ldots[ By (1)]$
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Question 852 Marks
Evaluate:
$\int \frac{3}{\sqrt{7 x-2}-\sqrt{7 x-5}} \cdot d x$
Answer
$\int \frac{3}{\sqrt{7 x-2}-\sqrt{7 x-5}} d x$
$=\int \frac{3}{\sqrt{7 x-2}-\sqrt{7 x-5}} \times \frac{\sqrt{7 x-2}+\sqrt{7 x-5}}{\sqrt{7 x-2}-\sqrt{7 x-5}} d x$
$=\int \frac{3(\sqrt{7 x-2}+\sqrt{7 x-5})}{(7 x-2)-(7 x-5)} d x$
$=\int(\sqrt{7 x-2}+\sqrt{7 x-5}) d x$
$=\int(7 x-2)^{\frac{1}{2}} d x+\int(7 x-5)^{\frac{1}{2}} d x$
$=\frac{(7 x-2)^{\frac{3}{2}}}{3 / 2} \times \frac{1}{7}+\frac{(7 x-5)^{\frac{3}{2}}}{3 / 2} \times \frac{1}{7}+c$
$=\frac{2}{21}(7 x-2)^{\frac{3}{2}}+\frac{2}{21}(7 x-5)^{\frac{2}{2}}+c .$
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Question 862 Marks
Evaluate:
$\int \frac{2}{\sqrt{x}-\sqrt{x+3}} \cdot d x$
Answer
$ \int \frac{2}{\sqrt{x}-\sqrt{x+3}} d x$
$=\int \frac{2}{\sqrt{x}-\sqrt{x+3}} \times \frac{\sqrt{x}+\sqrt{x+3}}{\sqrt{x}+\sqrt{x+3}} d x$
$=\int \frac{2(\sqrt{x}+\sqrt{x+3})}{x-(x+3)} d x$
$=-\frac{2}{3} \int(\sqrt{x}+\sqrt{x+3}) d x$
$=-\frac{2}{3} \int x^{\frac{1}{2}} d x-\frac{2}{3} \int(x+3)^{\frac{1}{2}} d x$
$=-\frac{2}{3} \cdot \frac{x^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}-\frac{2}{3} \cdot \frac{(x+3)^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}+c$
$=-\frac{4}{9}\left[x^{\frac{3}{2}}+(x+3)^{\frac{3}{2}}\right]+c .$
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Question 872 Marks
Evaluate:
$\int \cos ^2 x \cdot d x$
Answer
Recall the identity $\cos 2 x=2 \cos ^2 x-1$,
which gives$\cos ^2 x=\frac{1+\cos 2 x}{2}$
Therefore, $\int \cos ^2 x \cdot d x$
$=\frac{1}{2} \int(1+\cos 2 x) \cdot d x$
$=\frac{1}{2} \int d x+\frac{1}{2} \int \cos 2 x \cdot d x$
$=\frac{x}{2}+\frac{1}{4} \sin 2 x+C .$
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Question 882 Marks
Evaluate:

$\int \sqrt{1+\sin 5 x} \cdot d x$

Answer
$\begin{aligned} & \int \sqrt{1+\sin 5 x} \cdot d x \\ = & \int \sqrt{\sin ^2 x+\cos ^2 x+5 \sin x \cos x} \cdot d x \\ = & \int \sqrt{(\sin x+\cos x)^2} \cdot d x \\ = & \int(\sin x+\cos x) \cdot d x \\ = & \int \sin x d x+\int \cos x \cdot d x \\ = & \left(\frac{2}{5} \sin \frac{5 x}{2}-\cos \frac{5 x}{2}\right)+c .\end{aligned}$
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Question 892 Marks
Evaluate:
$\int \frac{\sin 4 x}{\cos 2 x} \cdot d x$
Answer
$\int \frac{\sin 4 x}{\cos 2 x} d x$
$=\int \frac{2 \sin 2 x \cos 2 x}{\cos 2 x} d x$
$=2 \int \sin 2 x d x$
$=2\left(-\frac{\cos 2 x}{2}\right)+c$
$=-\cos 2 x+c .$
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Question 902 Marks
Evaluate:
$\int \frac{2 x-7}{\sqrt{4 x-1}} \cdot d x$
Answer
$ \int \frac{2 x-7}{\sqrt{4 x-1}} d x$
$= \frac{1}{2} \int \frac{2(2 x-7)}{\sqrt{4 x-1}} d x$
$=  \frac{1}{2} \int \frac{(4 x-1)-13}{\sqrt{4 x-1}} d x$
$=\frac{1}{2} \int\left(\frac{4 x-1}{\sqrt{4 x-1}}-\frac{13}{\sqrt{4 x-1}}\right) d x$
$=\frac{1}{2} \int(4 x-1)^{\frac{1}{2}} d x-\frac{13}{2} \int(4 x-1)^{-\frac{1}{2}} d x$
$=\frac{1}{2} \cdot \frac{(4 x-1)^{\frac{3}{2}}}{(4)\left(\frac{3}{2}\right)}-\frac{13}{2} \cdot \frac{(4 x-1)^{\frac{1}{2}}}{(4)\left(\frac{1}{2}\right)}+c$
$=\frac{1}{12}(4 x-1)^{\frac{3}{2}}-\frac{13}{4} \sqrt{4 x-1}+c$.
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Question 912 Marks
Evaluate:

$\int \frac{x-2}{\sqrt{x+5}} \cdot d x$

Answer
$\begin{aligned} & \int \frac{x-2}{\sqrt{x+5}} d x=\int \frac{(x+5)-7}{\sqrt{x+5}} d x \\ = & \int\left(\frac{x+5}{\sqrt{x+5}}-\frac{7}{\sqrt{x+5}}\right) d x \\ = & \int(x+5)^{\frac{1}{2}} d x-7 \int(x+5)^{-\frac{1}{2}} d x \\ = & \frac{(x+5)^{\frac{3}{2}}}{(3 / 2)}-\frac{7(x+5)^{\frac{1}{2}}}{(1 / 2)}+c\end{aligned}$

$=\frac{2}{3}(x+5)^{\frac{3}{2}}-14 \sqrt{x+5}+c$.

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Question 922 Marks
Evaluate:
$\int \frac{5 x+2}{3 x-4} \cdot d x$
Answer
$\int \frac{5 x+2}{3 x-4} d x$
$=\int \frac{\frac{5}{3}(3 x-4)+\frac{20}{3}+2}{3 x-4} d x$
$=\int \frac{\frac{5}{3}(3 x-4)+\frac{26}{3}}{3 x-4} d x$
$=\int\left[\frac{5}{3}+\frac{\left(\frac{26}{3}\right)}{3 x-4}\right] d x$
$=\frac{5}{3} \int 1 d x+\frac{26}{3} \int \frac{1}{3 x-4} d x$
$=\frac{5 x}{3}+\frac{26}{3} \cdot \frac{1}{3} \log |3 x-4|+c$
$=\frac{5 x}{3}+\frac{26}{9} \log |3 x-4|+c .$
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Question 932 Marks
Evaluate:
$\int \frac{4 x+3}{2 x+1} \cdot d x$
Answer
$ \int \frac{4 x+3}{2 x+1} d x $
$ =  \int \frac{2(2 x+1)+1}{2 x+1} d x$
$=\int\left(\frac{2(2 x+1)}{2 x+1}+\frac{1}{2 x+1}\right) d x$
$=2 \int 1 d x+\int \frac{1}{2 x+1} d x$
$=2 x+\frac{1}{2} \log |2 x+1|+c .$
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Question 942 Marks
Evaluate:
$\int \frac{x}{x+2} \cdot d x$
Answer
$\quad \int \frac{x}{x+2} d x$
$=\int \frac{(x+2)-2}{x+2} d x$
$=\int\left(\frac{x+2}{x+2}-\frac{2}{x+2}\right) d x$
$=\int 1 d x-2 \int \frac{1}{x+2} d x$
$=x-2 \log |x+2|+c .$
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Solve the Following Question.(2 Marks) - Page 2 - Maths STD 12 Questions - Vidyadip