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Maths Solve the Following Question.(4 Marks)

Indefinite Integration · Maharashtra Board STD 12 (MSBSHSE - Semi English Medium). 26 questions.

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Solve the Following Question.(4 Marks)

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26 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks
Integrate the following with respect to the respective variable:

$\cot ^{-1}\left(\frac{1+\sin x}{\cos x}\right)$

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Question 84 Marks
Evaluate : $\int \frac{3 x^2+4 x-5}{\left(x^2-1\right)(x+2)} d x$
Answer
$\mathrm{I}=\int \frac{3 x^2+4 x-5}{(x-1)(x+1)(x+2)} \cdot d x$
$
\begin{aligned}
\text { Consider, } \quad \frac{3 x^2+4 x-5}{(x-1)(x+1)(x+2)} & =\frac{A}{(x-1)}+\frac{B}{(x+1)}+\frac{C}{(x+2)} \\
& =\frac{A(x+1)(x+2)+B(x-1)(x+2)+C(x-1)(x+1)}{(x-1)(x+1)(x+2)} \\
\therefore \quad 3 x^2+4 x-5=A(x+1)(x+2) & +B(x-1)(x+2)+C(x-1)(x+1) \\
\text { at } x=1, \quad 3(1)^2+4(1)-5 & =A(2)(3)+B(0)+C(0) \\
2=6 A \quad \Rightarrow A & =\frac{1}{3} \\
\text { at } x=-1, \quad 3(-1)^2+4(-1)-5 & =A(0)+B(-2)(1)+C(0) \\
-6=-2 B \Rightarrow B & =3
\end{aligned}
$
at $x=1$,
$
3(1)^2+4(1)-5=A(2)(3)+B(0)+C(0)
$
at $x=-1$,
$
3(-1)^2+4(-1)-5=A(0)+B(-2)(1)+C(0)
$
at $x=-2, \quad 3(-2)^2+4(-2)-5=A(0)+B(0)+C(-3)(-1)$
$
-1=3 C \quad \Rightarrow \quad C=-\frac{1}{3}
$
Thus, $\quad \frac{3 x^2+4 x-5}{(x-1)(x+1)(x+2)}=\frac{\left(\frac{1}{3}\right)}{(x-1)}+\frac{3}{(x+1)}+\frac{\left(-\frac{1}{3}\right)}{(x+2)}$
$
\begin{aligned}
\therefore \mathrm{I} & =\int\left[\frac{\left(\frac{1}{3}\right)}{(x-1)}+\frac{3}{(x+1)}+\frac{\left(-\frac{1}{3}\right)}{(x+2)}\right] \cdot d x & =\frac{1}{3} \log (x-1)+3 \log (x+1)-\frac{1}{3} \log (x+2)+c \\
& =\frac{1}{3} \log \left[\frac{(x-1)(x+1)^9}{(x+2)}\right]+c & \therefore \int \frac{3 x^2+4 x-5}{\left(x^2-1\right)(x+2)} \cdot d x=\frac{1}{3} \log \left[\frac{(x-1)(x+1)^9}{(x+2)}\right]+c
\end{aligned}
$
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Question 94 Marks
Evaluate : $\int \sqrt{4+3 x-2 x^2} \cdot d x$
Answer
$
\begin{gathered}
\mathrm{I}=\int \sqrt{4-2 x^2+3 x} \cdot d x \\
=\int \sqrt{4-2\left(x^2-\frac{3}{2} x\right)} \cdot d x \\
=\int \sqrt{2} \cdot \sqrt{2-\left(x^2-\frac{3}{2} x\right)} \cdot d x
\end{gathered}
$
$
\because \quad\left\{\left(\frac{1}{2} \text { coefficient of } x\right)^2=\left[\frac{1}{2}\left(-\frac{3}{2}\right)\right]^2=\left(-\frac{3}{4}\right)^2=\frac{9}{16}\right\}
$
$
\begin{aligned}
& \mathrm{I}=\sqrt{2} \cdot \int \sqrt{2-\left(x^2-\frac{3}{2} x+\frac{9}{16}-\frac{9}{16}\right)} \cdot d x \\
& =\sqrt{2} \cdot \int \sqrt{2-\left(x^2-\frac{3}{2} x+\frac{9}{16}\right)+\frac{9}{16}} \cdot d x \\
& =\sqrt{2} \cdot \int \sqrt{\left(\frac{\sqrt{41}}{4}\right)^2-\left(x-\frac{3}{4}\right)^2} \cdot d x \\
& \because \int \sqrt{a^2-x^2} \cdot d x=\frac{x}{2} \cdot \sqrt{a^2-x^2}+\frac{a^2}{2} \cdot \sin ^{-1}\left(\frac{x}{a}\right)+c \\
& =\sqrt{2} \cdot\left[\frac{\left(x-\frac{3}{4}\right)}{2} \cdot \sqrt{\left(\frac{\sqrt{41}}{4}\right)^2-\left(x-\frac{3}{4}\right)^2}+\frac{\left(\frac{\sqrt{41}}{4}\right)^2}{2} \cdot \sin ^{-1}\left(\frac{x-\frac{3}{4}}{\frac{\sqrt{41}}{4}}\right)\right]+c \\
& =\sqrt{2}\left[\frac{4 x-3}{8} \cdot \sqrt{2+\frac{3}{2} x-x^2}+\frac{41}{32} \cdot \sin ^{-1}\left(\frac{4 x-3}{\sqrt{41}}\right)\right]+c \\
& \therefore \int \sqrt{4+3 x-2 x^2} \cdot d x=\frac{4 x-3}{8} \cdot \sqrt{4+3 x-2 x^2}+\frac{41}{16 \sqrt{2}} \cdot \sin ^{-1}\left(\frac{4 x-3}{\sqrt{41}}\right)+c \\
&
\end{aligned}
$
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Question 104 Marks
Evaluate : $\int \sqrt{x^2-a^2} \cdot d x=\frac{x}{2} \cdot \sqrt{x^2-a^2}-\frac{a^2}{2} \cdot \log \left(x+\sqrt{x^2+a^2}\right)+c$
Answer
Self
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Question 114 Marks
Evaluate : $\int \sqrt{a^2+x^2} \cdot d x=\frac{x}{2} \cdot \sqrt{x^2+a^2}+\frac{a^2}{2} \cdot \log \left(x+\sqrt{x^2+a^2}\right)+c$
Answer
Self
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Question 124 Marks
Evaluate : $\int \sqrt{a^2-x^2} \cdot d x$
Answer
Let $\mathrm{I}=\int \sqrt{a^2-x^2} \cdot 1 \cdot d x$
$
\begin{aligned}
& =\sqrt{a^2-x^2} \cdot \int 1 \cdot d x-\int \frac{d}{d x} \cdot \sqrt{a^2-x^2} \cdot \int 1 \cdot d x \cdot d x \\
& =\sqrt{a^2-x^2} \cdot x-\int \frac{1}{2 \sqrt{a^2-x^2}}(-2 x) \cdot(x) \cdot d x \\
& =\sqrt{a^2-x^2} \cdot x+\int \frac{x^2}{\sqrt{a^2-x^2}} \cdot d x \\
& =\sqrt{a^2-x^2} \cdot x+\int \frac{a^2-\left(a^2-x^2\right)}{\sqrt{a^2-x^2}} \cdot d x \\
& =\sqrt{a^2-x^2} \cdot x+\int\left[\frac{a^2}{\sqrt{a^2-x^2}}-\frac{\left(a^2-x^2\right)}{\sqrt{a^2-x^2}}\right] \cdot d x \\
& =x \cdot \sqrt{a^2-x^2}+a^2 \int \frac{1}{\sqrt{a^2-x^2}} \cdot d x-\int \sqrt{a^2-x^2} \cdot d x \\
& \mathrm{I}=x \cdot \sqrt{a^2-x^2}+a^2 \int \frac{1}{\sqrt{a^2-x^2}} \cdot d x-\mathrm{I} \\
& \therefore \quad \mathrm{I}+\mathrm{I} \quad=x \cdot \sqrt{a^2-x^2}+a^2 \cdot \sin ^{-1}\left(\frac{x}{a}\right)+c \\
& \therefore \quad \text { I } \quad=\frac{x}{2} \cdot \sqrt{a^2-x^2}+\frac{a^2}{2} \cdot \sin ^{-1}\left(\frac{x}{a}\right)+c \\
& \therefore \quad \int \sqrt{a^2-x^2} \cdot d x=\frac{x}{2} \cdot \sqrt{a^2-x^2}+\frac{a^2}{2} \cdot \sin ^{-1}\left(\frac{x}{a}\right)+c \\
& \text { e.g. } \int \sqrt{9-x^2} \cdot d x=\frac{x}{2} \cdot \sqrt{9-x^2}+\frac{9}{2} \cdot \sin ^{-1}\left(\frac{x}{3}\right)+c \\
&
\end{aligned}
$
with reference to the above example solve these :
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Question 134 Marks
Evaluate : $\int\left[\log (\log x)+\frac{1}{(\log x)^2}\right] \cdot d x $
Answer
$\begin{aligned}\text {I } \quad=\int \log (\log x) \cdot 1 \cdot d x+\int \frac{1}{(\log x)^2} \cdot d x \\ & =\log (\log x) \cdot \int 1 \cdot d x-\int \frac{d}{d x} \cdot \log (\log x) \int 1 \cdot d x+\int \frac{1}{(\log x)^2} \cdot d x \\ & =\log (\log x) \cdot x-\int \frac{1}{\log x} \cdot \frac{1}{x} \cdot(x) \cdot d x+\int \frac{1}{(\log x)^2} \cdot d x \\ & =\log (\log x) \cdot x-\int \frac{1}{\log x} \cdot d x+\int \frac{1}{(\log x)^2} \cdot d x \\ & =\log (\log x) \cdot x-\int(\log x)^{-1} \cdot 1 \cdot d x+\int \frac{1}{(\log x)^2} \cdot d x \\ & =\log (\log x) \cdot x-\left\{(\log x)^{-1} \cdot \int 1 \cdot d x+\int \frac{d}{d x} \cdot(\log x)^{-1} \cdot \int 1 \cdot d x \cdot d x\right\}+\int \frac{1}{(\log x)^2} \cdot d x \\ & =\log (\log x) \cdot x-\left\{(\log x)^{-1} \cdot x-\int-1(\log x)^{-2} \cdot \frac{1}{x} \cdot x \cdot d x\right\}+\int \frac{1}{(\log x)^2} \cdot d x \\ & =\log (\log x) \cdot x-(\log x)^{-1} \cdot x-\int(\log x)^{-2} \cdot d x+\int \frac{1}{(\log x)^2} \cdot d x \\ & =x \cdot \log (\log x)-\frac{x}{\log x}-\int \frac{1}{(\log x)^2} \cdot d x+\int \frac{1}{(\log x)^2} \cdot d x \\ & \therefore \int\left[\log (\log x)+\frac{1}{(\log x)^2}\right] \cdot d x=x \cdot \log (\log x)-\frac{x}{\log x}+c \\ & \end{aligned}$
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Question 144 Marks
Evaluate : $\int e^{2 x} \cdot \sin 3 x \cdot d x$
Answer
$\quad \mathrm{I}=\int e^{2 x \cdot} \sin 3 x \cdot d x$
Here we use repeated integration by parts.
To evaluate $\int e^{a x} \cdot \sin (b x+c) \cdot d x ; \int e^{a x} \cdot \cos (b x+c) \cdot d x$ any function can be taken as a first function.
$
\begin{aligned}
& \mathrm{I}=e^{2 x} \cdot \int \sin 3 x \cdot d x-\int \frac{d}{d x} \cdot e^{2 x} \cdot \int \sin 3 x \cdot d x \cdot d x \\
& =e^{2 x} \cdot\left(-\cos 3 x \cdot \frac{1}{3}\right)-\int e^{2 x \cdot 2}\left(-\cos 3 x \cdot \frac{1}{3}\right) \cdot d x \\
& =-\frac{1}{3} \cdot e^{2 x} \cdot \cos 3 x+\frac{2}{3} \int e^{2 x} \cdot \cos 3 x \cdot d x \\
& =-\frac{1}{3} \cdot e^{2 x} \cdot \cos 3 x+\frac{2}{3}\left(e^{2 x} \cdot \int \cos 3 x \cdot d x-\int \frac{d}{d x} \cdot e^{2 x} \cdot \int \cos 3 x \cdot d x \cdot d x\right) \\
& =-\frac{1}{3} \cdot e^{2 x} \cdot \cos 3 x+\frac{2}{3}\left[e^{2 x \cdot}\left(\sin 3 x \cdot \frac{1}{3}\right)-\int e^{2 x \cdot 2} \cdot\left(\sin 3 x \cdot \frac{1}{3}\right) \cdot d x\right] \\
& =-\frac{1}{3} \cdot e^{2 x} \cdot \cos 3 x+\frac{2}{9} \cdot e^{2 x} \cdot \sin 3 x-\frac{4}{9} \cdot \int e^{2 x} \cdot \sin 3 x \cdot d x \\
& I=-\frac{1}{3} \cdot e^{2 x} \cdot \cos 3 x+\frac{2}{9} \cdot e^{2 x} \cdot \sin 3 x-\frac{4}{9} \cdot I \\
\end{aligned}
$
$\begin{aligned} & \mathrm{I}+\frac{4}{9} \cdot \mathrm{I}=\frac{e^{2 x}}{9}[-3 \cos 3 x+2 \sin 3 x]+c \\ & \frac{13}{9} \cdot \mathrm{I}=\frac{e^{2 x}}{9}[2 \sin 3 x-3 \cos 3 x]+c\end{aligned}$
$\begin{gathered}=\frac{e^{2 x}}{13}[2 \sin 3 x-3 \cos 3 x]+c \\ \therefore \int e^{2 x \cdot} \sin 3 x \cdot d x=\frac{e^{2 x}}{13}[2 \sin 3 x-3 \cos 3 x]+c\end{gathered}$
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Question 154 Marks
Evaluate : $\int \frac{1}{5-4 \cos x} \cdot d x$
Answer
put $\tan \frac{x}{2}=t$
$
\begin{aligned}
& \therefore \quad d x=\frac{2}{1+t^2} \cdot d t \text { and } \quad \cos x=\frac{1-t^2}{1+t^2} \\
& I=\int \frac{1\left(\frac{2}{1+t^2}\right)}{5-4\left(\frac{1-t^2}{1+t^2}\right)} \cdot d t \\
& =\int \frac{\frac{2}{1+t^2}}{\frac{5\left(1+t^2\right)-4\left(1-t^2\right)}{1+t^2}} \cdot d t \\
& =\int \frac{2}{5-5 t^2-4-4 t^2} \cdot d t \\
& =\int \frac{2}{9 t^2+1} \cdot d t \\
& =\int \frac{1}{9\left(t^2+\frac{1}{9}\right)} \cdot d t \\
& =\frac{2}{9} \cdot \int \frac{1}{t^2+\left(\frac{1}{3}\right)^2} \cdot d t \\
& =\frac{2}{9} \cdot \frac{1}{\left(\frac{1}{3}\right)} \cdot \tan ^{-1}\left(\frac{t}{\left(\frac{1}{3}\right)}\right)+c \\
& =\frac{2}{3} \cdot \tan ^{-1}(2 t)+c \\
& =\frac{2}{3} \cdot \tan ^{-1}\left(2 \tan \frac{x}{2}\right)+c \\
& \therefore \int \frac{1}{5-4 \cos x} \cdot d x=\frac{2}{3} \cdot \tan ^{-1}\left(2 \tan \frac{x}{2}\right)+c \\
\end{aligned}
$
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