4 Marks

Question 14 Marks

Evaluate $\int \frac{x+1}{x^2+4 x+5} d x$.

Answer

Let
$
\begin{aligned}
I & =\int \frac{x+1}{x^2+4 x+5} d x \\
x+1 & =A \frac{d}{d x}\left(x^2+4 x+5\right)+B \\
x+1 & =A(2 x+4)+B
\end{aligned}
$
On comparing coefficients of $x$ and constants, Now
$
\begin{aligned}
2 A & =1 & \therefore A=\frac{1}{2} \\
I & =4 A+B & \\
I & =4 \times \frac{1}{2}+B & \therefore B=1-2=-1
\end{aligned}
$
now$\quad x+1=\frac{1}{2}(2 x+4)-1$
$
\begin{array}{l}
\therefore \int \frac{x+1}{x^2+4 x+5} d x=\int \frac{\frac{1}{2}(2 x+4)-1}{x^2+4 x+5} d x \\
=\frac{1}{2} \int \frac{2 x+4}{x^2+4 x+5} d x-\int \frac{1}{x^2+4 x+5} d x
\end{array}
$
For $I _1$ :
Let
$
\begin{aligned}
x^2+4 x+5 & =t \\
(2 x+4) d x & =d t \\
I_1 & =\frac{1}{2} \int \frac{d t}{t}=\frac{1}{2} \log |t|+C_1 \\
& =\frac{1}{2} \log \left|x^2+4 x+5\right|+C_1
\end{aligned}
$
For $I _2$ :
$
\int \frac{1}{x^2+4 x+5} d x
$
Here
$
\begin{aligned}
x^2+4 x+5 & =x^2+4 x+2^2-2^2+5 \\
& =(x+2)^2+1 \\
& =\int \frac{1}{(x+2)^2+1} d x=\int \frac{1}{(x+2)^2+(1)^2} d x
\end{aligned}
$
Let $\quad x+2=t$ then $d x=d t$
$
\begin{aligned}
I_2 & =\int \frac{d t}{t^2+(1)^2}=\tan ^{-1} t+C_2 \\
& =\tan ^{-1}(x+2)+C_2
\end{aligned}
$
Putting values of $I _1$ and $I _2$
$
I=\left\{\frac{1}{2} \log \left|x^2+4 x+5\right|+C_1\right\}-\left\{\tan ^{-1}(x+2)+C_2\right\}
$
On writing new constant C in place of $C _1$ and $C _2$ $I =\frac{1}{2} \log \left|x^2+4 x+5\right|-\tan ^{-1}(x+2)+C$ Ans.

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Question 24 Marks

Evaluate $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \sin ^2 x d x$.

Answer

Here $\sin ^2 x$ is an even function.
So, $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \sin ^2 x d x=2 \int_0^{\frac{\pi}{4}} \sin ^2 x d x$
$
\left(\because \int_{-a}^a f(x) d x=2 \int_0^a f(x) d x\right)
$
$\begin{array}{l}=2 \int_0^{\frac{\pi}{4}} \frac{1-\cos 2 x}{2} d x \\ =2 \int_0^{\frac{\pi}{4}}(1-\cos 2 x) d x \\ =\left(x-\frac{1}{2} \sin 2 x\right)_0^{\frac{\pi}{4}} \\ =\left(\frac{\pi}{4}-\frac{1}{2} \sin \frac{\pi}{2}\right)-0=\frac{\pi}{4}-\frac{1}{2} \text { }\end{array}$

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Question 34 Marks

Evaluate $\int_{-1}^1 5 x^4 \sqrt{x^5+1} d x$.

Answer

Let
$
I=\int_{-1}^1 5 x^4 \sqrt{x^5+1} d x
$
Putting $t=x^5+1 \Rightarrow d t=5 x^4 d x$
So,
$
\begin{aligned}
\int 5 x^4 \sqrt{x^5+1} d x & =\int \sqrt{t} d t \\
= & \frac{2}{3} t^{\frac{3}{2}}=\frac{2}{3}\left(x^5+1\right)^{\frac{3}{2}}
\end{aligned}
$
So, $\int_{-1}^1 5 x^4 \sqrt{x^5+1} d x$
$
\begin{array}{l}
=\frac{2}{3}\left(\left(x^5+1\right)^{\frac{3}{2}}\right)_{-1}^1 \\
=\frac{2}{3}\left(\left(1^5+1\right)^{\frac{3}{2}}-\left((-1)^5+1\right)^{\frac{3}{2}}\right) \\
=\frac{2}{3}\left(2^{\frac{3}{2}}-(0)^{\frac{3}{2}}\right)=\frac{2}{3}(2 \sqrt{2}-0) \\
=\frac{4 \sqrt{2}}{3} \text { }
\end{array}
$

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Question 44 Marks

Evaluate :
$
\int_0^{\frac{\pi}{4}}\left(\frac{\sin x+\cos x}{3+\sin 2 x}\right) d x
$

Answer

Let
$
I=\int_0^{\frac{\pi}{4}} \frac{\sin x+\cos x}{3+\sin 2 x} d x
$
Here putting $\sin x-\cos x= t$
$
(\cos x+\sin x) d x=d t
$
So if $x=\frac{\pi}{4}$ then $t=0$ and if $x=0$ then $t=-1$
$
\begin{array}{l}
\begin{array}{l}
(\sin x-\cos x)^2=1-2 \sin x \cos x=1-\sin 2 x \\
\therefore \sin 2 x=1-(\sin x-\cos x)^2 \\

=1-t^2
\end{array} \\
\begin{aligned}
I & =\int_{-1}^0 \frac{d x}{3+\left(1-t^2\right)}=\int_{-1}^0 \frac{d t}{4-t^2}
\end{aligned} \\
I=\int_{-1}^0 \frac{d t}{(2)^2-t^2}=\frac{1}{2 \times 2}\left(\log \frac{2+t}{2-t}\right)_{-1}^0 \\
\quad\left(\because \int \frac{1}{a^2-x^2} d x=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+c\right)
\end{array}
$
$
\begin{array}{l}
I=\frac{1}{4}\left(\log 1-\log \frac{1}{3}\right) \\
I=\frac{1}{4}\left(0-\log \frac{1}{3}\right)=\frac{1}{4} \log 3
\end{array}
$

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Question 54 Marks

Evaluate :
$
\int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x
$

Answer

<$
\begin{aligned}
I & =\int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x \\
& =\int_0^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-x\right) \sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)}{\sin ^4\left(\frac{\pi}{2}-x\right)+\cos ^4\left(\frac{\pi}{2}-x\right)} d x \\
& {\left(\because \int_0^a f(x) d x=\int_0^a f(a-x) d x\right) } \\
& \int_0^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-x\right) \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x
\end{aligned}
$
Adding eqns (1) and (2)
$
\begin{aligned}
2 I & =\frac{\pi}{2} \int_0^{\frac{\pi}{2}} \frac{\cos x \sin x}{\sin ^4 x+\cos ^4 x} d x \\
\Rightarrow \quad I & =\frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{\cos x \sin x}{\sin ^4 x+\cos ^4 x} d x \\
& =\frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{\tan x \sec ^2 x}{1+\tan ^4 x} d x
\end{aligned}
$
$\begin{array}{l}\text { Putting } \tan ^2 x=t \Rightarrow 2 \tan x \sec ^2 x d x=d t \\ \qquad \begin{aligned} \therefore \quad I & =\frac{\pi}{8} \int_0^{\infty} \frac{1}{1+t^2} d t \\ & \quad\left(\because x=0 \text { then } t=0 ; x=\frac{\pi}{2} \text { then } t=\infty\right) \\ & =\frac{\pi}{8}\left(\tan ^{-1} t\right)_0^{\infty} \\ & =\frac{\pi}{8}\left(\tan ^{-1}(\infty)-\tan ^{-1}(0)\right) \\ & =\frac{\pi}{8}\left(\frac{\pi}{2}-0\right)\end{aligned}\end{array}$
$=\frac{\pi^2}{16}$

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Question 64 Marks

Integrate the function $\frac{e^{a \tan ^{-1} x}}{\left(1+x^2\right)^{3 / 2}}$ with respect to $x$.

Answer

$
\begin{array}{l}
\text { Let } \\
I=\int \frac{e^{a \tan ^{-1} x}}{\left(1+x^2\right)^{3 / 2}} d x \\
\text { Putting } \quad \tan ^{-1} x=t \\
\therefore \quad x=\tan t \\
\therefore \frac{1}{1+x^2} d x=d t \\
I=\int \frac{e^{a \tan ^{-1} x}}{\sqrt{1+x^2}\left(1+x^2\right)} d x \\
=\int \frac{e^{a t} d t}{\sqrt{1+\tan ^2 t}}=\int \frac{e^{a t} d t}{\sec t} \\
I=\iint_{II}^{a t} \underset{I}{\cos t} d t \\
\text { (using ILATE) } \\
I=\cos t \cdot \frac{e^{a t}}{a}-\int \frac{(-\sin t) \cdot e^{a t}}{a} d t \\
=\frac{1}{a} e^{a t} \cos t+\frac{1}{a} \int_{II} e^{a t} \sin _{I} t d t
\end{array}
$
$\begin{array}{l}=\frac{1}{a} e^{a t} \cos t+ \frac{1}{a}\left(\frac{\sin t \cdot e^{a t}}{a}-\int \cos t \cdot \frac{e^{a t}}{a} d t\right) \\ =\frac{1}{a} e^{a t} \cos t+\frac{1}{a^2} e^{a t} \sin t-\frac{1}{a^2} \int e^{a t} \cos t d t \\ =\frac{1}{a^2} e^{a t}(a \cos t+\sin t)-\frac{1}{a^2} I \\ \left(1+\frac{1}{a^2}\right) I =\frac{e^{a t}}{a^2}(a \cos t+\sin t)+ C \\ I =\frac{e^{a t}}{a^2} \times \frac{a^2}{\left(a^2+1\right)}(a \cos t+\sin t)+ C \\ = \frac{e^{a t}}{\left(a^2+1\right)}(a \cos t+\sin t)+ C \\ \therefore \tan -1 x=t \Rightarrow x=\tan t \\ = \frac{e^{a \tan ^{-1} x}}{a^2+1}\left(a \cdot \frac{1}{\sqrt{1+x^2}}+\frac{x}{\sqrt{1+x^2}}\right)+ C \\ = \frac{e^{a \tan ^{-1} x}}{\left(1+a^2\right)} \cdot \frac{(a+x)}{\sqrt{1+x^2}}+ C \quad Ans .\end{array}$

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Question 74 Marks

Evaluate :
$
\int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x
$

Answer

Let
$
\begin{array}{l}
I=\int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x \\
I=\int_{-\pi}^\pi \frac{2 x+2 x \sin x}{1+\cos ^2 x} d x
\end{array}
$
$
\begin{array}{l}
I=\int_{-\pi}^\pi \frac{2 x d x}{1+\cos ^2 x}+\int_{-\pi}^\pi \frac{2 x \sin x}{1+\cos ^2 x} d x \\
I=I_1+I_2\quad \quad \ldots \ldots(1)
\end{array}
$
Here $_1=\int_{-\pi}^\pi \frac{2 x d x}{1+\cos ^2 x}$ and $I _2=\int_{-\pi}^\pi \frac{2 x \sin x}{1+\cos ^2 x} d x$
$\therefore \frac{2 x}{1+\cos ^2 x}$ is an odd function and $\frac{2 x \sin x}{1+\cos ^2 x}$ is an even function.
$\therefore \quad I _1=0$ and $I _2=2 \int_0^\pi \frac{2 x \sin x}{1+\cos ^2 x} d x$
Now $\quad I _2=4 \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x\quad \quad \ldots \ldots(2)$
Using property $P _5$ in $I _2$,
$\Rightarrow I _2=4 \int_0^\pi \frac{(\pi-x) \sin (\pi-x)}{1+\cos ^2(\pi-x)} d x$
$
\left(\because \int_0^a f(x) d x=\int_0^a f(a-x) d x\right)
$
$
\begin{array}{l}
I_2=4 \int_0^\pi \frac{(\pi-x) \sin x}{1+\cos ^2 x} d x \\
I_2=4 \int_0^\pi \frac{\pi \sin x}{1+\cos ^2 x} d x-4 \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x \ldots \\
I_2=4 \pi \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x-I_2 \\
2 I_2=4 \pi \int_0^\pi \frac{\sin x d x}{1+\cos ^2 x} \\
\text { putting } \begin{aligned}
\cos x=t \Rightarrow-\sin x d x=d t \\
\sin x d x=-d t
\end{aligned}
\end{array}
$
0 n changing limits when $x=0$, then $t=1$ and when $x=$ then $t=-1$
$
\begin{aligned}
2 I_2 & =4 \pi \int_1^{-1} \frac{-d t}{1+t^2} \\
2 I_2 & =4 \pi \int_{-1}^1 \frac{d t}{1+t^2} \\
I_2 & =2 \pi\left(\tan ^{-1} t\right)_{-1}^1 \\
& =2 \pi\left(\tan ^{-1} 1-\tan ^{-1}(-1)\right) \\
& =2 \pi\left(\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right)
\end{aligned}
$ (using property $P _2$ )
$
\begin{aligned}
& =2 \pi\left(\frac{\pi}{4}+\frac{\pi}{4}\right)=2 \pi \times \frac{2 \pi}{4}=\frac{4 \pi^2}{4} \\
I_2 & =\pi^2
\end{aligned}
$
Putting $I _1$ and $I _2$ in eq. (1)
$
I=0+\pi^2=\pi^2
$
$\Rightarrow \int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x=\pi^2$

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Question 84 Marks

Evaluate $\int e^x\left(\frac{1-x}{1+x}\right)^2 d x$

Answer

Let
$
\begin{aligned}
I & =\int e^x\left(\frac{1-x}{1+x}\right)^2 d x \\
I & =\int \frac{e^x(1-x)^2}{(1+x)^2} d x=\int e^x\left(\frac{(1+x)^2-4 x}{(1+x)^2}\right) d x \\
& =\int e^x d x-4 \int \frac{x e^x}{(1+x)^2} d x \\
& =\int e^x d x-4 \int e^x \frac{(x+1-1)}{(1+x)^2} d x \\
& =\int e^x d x-4 \int\left(\frac{1}{(1+x)}-\frac{1}{(1+x)^2}\right) e^x d x \\
& =e^x-4 \int\left(\frac{1}{1+x}-\frac{1}{(1+x)^2}\right) e^x d x
\end{aligned}
$
Let $\quad f(x)=\frac{1}{1+x} \therefore f^{\prime}(x)=\frac{-1}{(1+x)^2}$
So it is in the form of $\int\left(f(x)+f^{\prime}(x)\right) e^x d x$.
So its value is equal to $e^x f(x)+C$.
$
\begin{aligned}
I & =e^x-4\left(e^x \times \frac{1}{1+x}\right)+C \\
& \quad\left(\because f(x)=\frac{1}{1+x}\right) \\
& =e^x\left(\frac{1+x-4}{1+x}\right)+C=\left(\frac{x-3}{x+1}\right) e^x+C \text { Ans. }
\end{aligned}
$

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Question 94 Marks

Evaluate $\int \frac{x^2 d x}{(x \sin x+\cos x)^2}$.

Answer

Let
$
I=\int \frac{x^2 d x}{(x \sin x+\cos x)^2}
$
Since $\frac{d}{d x}(x \sin x+\cos x)=1 \cdot \sin x+x \cos x-\sin x$
$
=x \cos x
$
So assuming $x^2$ as $\frac{x}{\cos x} \cdot x \cos x$ taking
$\frac{x \cos x}{(x \sin x+\cos x)^2}$ as the second function, integrating by parts,
$
\begin{array}{l}
\int \frac{x^2}{(x \sin x+\cos x)^2} d x \\
=\int \frac{x}{\cos x} \cdot \frac{x \cos x}{(x \sin x+\cos x)^2} d x \\
=\frac{x}{\cos x} \times\left(\frac{-1}{x \sin x+\cos x}\right)- \\
\int \frac{\cos x+x \sin x}{\cos ^2 x} \times\left\{\frac{-1}{x \sin x+\cos x}\right\} d x \\
=\frac{-x}{\cos x(x \sin x+\cos x)}+\int \sec ^2 x d x
\end{array}
$
$\begin{array}{l}=\frac{-x}{\cos x(x \sin x+\cos x)}+\tan x+ C \\ =\frac{-x}{\cos x(x \sin x+\cos x)}+\frac{\sin x}{\cos x}+ C \\ =\frac{-x+x \sin ^2 x+\sin x \cos x}{\cos x(x \sin x+\cos x)}+ C \end{array}$
$\begin{array}{l}=\frac{-x \cos ^2 x+\sin x \cos x}{\cos x(x \sin x+\cos x)}+ C \\ =\frac{\sin x-x \cos x}{x \sin x+\cos x}+ C \text { }\end{array}$

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Question 104 Marks

Evaluate :
$
\int_0^{\frac{\pi}{2}} \frac{\cos ^2 x}{1+3 \sin ^2 x} d x
$

Answer

Let
$
\begin{aligned}
I & =\int_0^{\frac{\pi}{2}} \frac{\cos ^2 x d x}{1+3 \sin ^2 x}=\int_0^{\frac{\pi}{2}} \frac{\cos ^2 x}{\sin ^2 x+\cos ^2 x+3 \sin ^2 x} d x \\
I & =\int_0^{\frac{\pi}{2}} \frac{\cos ^2 x}{\cos ^2 x+4 \sin ^2 x} d x=\int_0^{\frac{\pi}{2}} \frac{\cos ^2 x}{\cos ^2 x+4\left(1-\cos ^2 x\right)} d x \\
I & =\int_0^{\frac{\pi}{2}} \frac{\cos ^2 x}{4-3 \cos ^2 x} d x \\
& =\frac{-1}{3} \int_0^{\frac{\pi}{2}} \frac{-3 \cos ^2 x}{4-3 \cos ^2 x} d x=\frac{-1}{3} \int_0^{\frac{\pi}{2}} \frac{4-3 \cos ^2 x-4}{4-3 \cos ^2 x} d x \\
& =\int_0^{\frac{\pi}{2}}\left(-\frac{1}{3}+\frac{4}{3\left(4-3 \cos ^2 x\right)}\right) d x \\
& =\frac{-1}{3}(x)_0^{\frac{\pi}{2}}+\frac{4}{3} \int_0^{\frac{\pi}{2}} \frac{d x}{4-3 \cos ^2 x} \\
& =-\frac{\pi}{6}+\frac{4}{3} \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{4 \sec ^2 x-3} d x
\end{aligned}
$
(On dividing by $\cos ^2 x$ in numerator and denominator)
$
=-\frac{\pi}{6}+\frac{4}{3} \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{1+4 \tan ^2 x} d x
$
Again putting $\tan x=t \Rightarrow \sec ^2 x d x=d t$,
If $x=\frac{\pi}{2}$ then $t=\tan \frac{\pi}{2}=\infty$, If $x=0$ then $t=0$
$
\begin{aligned}
I & =-\frac{\pi}{6}+\frac{4}{3} \int_0^{\infty} \frac{1}{1+4 t^2} d t \\
& =-\frac{\pi}{6}+\frac{4}{3} \times \frac{1}{2}\left(\tan ^{-1} 2 t\right)_0^{\infty} \\
I & =-\frac{\pi}{6}+\frac{2}{3} \times \frac{\pi}{2}=-\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{6} \text { Ans. }
\end{aligned}
$

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Question 114 Marks

Evaluate :
$
\int \frac{x^3}{(x-1)\left(x^2+1\right)} d x
$

Answer

$
\begin{array}{l}
\int \frac{x^3}{(x-1)\left(x^2-1\right)} d x \\
\Rightarrow \int \frac{x^3}{x^3-x^2+x-1} d x
\end{array}
$
Here degree of numerator and denominator is same. So on dividing numerator by denominator :
$
\begin{array}{l}
\int 1+\frac{x^2-x+1}{(x-1)\left(x^2+1\right)} d x \\
\int d x+\int \frac{x^2-x+1}{(x-1)\left(x^2+1\right)} d x
\end{array}
$
Using Partial fraction,
$
\begin{array}{l}
\frac{x^2-x+1}{(x-1)\left(x^2+1\right)}=\frac{A}{(x-1)}+\frac{B x+C}{x^2+1} \\
\Rightarrow x^2-x+1=A\left(x^2+1\right)+(B x+C)(x-1)
\end{array}
$
Putting $x=1 \Rightarrow 1=2 A \quad \therefore A =\frac{1}{2}$
Comparing coefficient of $x^2 \Rightarrow 1= A + B \therefore B =1-\frac{1}{2}$ $=\frac{1}{2}$
Comparing coefficient of $x-1=- B + C \therefore C =-\frac{1}{2}$
$
1+\frac{1}{2(x-1)}+\frac{\frac{1}{2} x-\frac{1}{2}}{x^2+1}
$
$
1+\frac{1}{2(x-1)}+\frac{1}{2} \frac{x}{x^2+1}-\frac{1}{2} \cdot \frac{1}{x^2+1}
$
On integrating,
$
\begin{array}{l}
\int 1 d x+\frac{1}{2} \int \frac{1}{(x-1)}+\frac{1}{2} \int \frac{x}{x^2+1} d x-\frac{1}{2} \int \frac{1}{x^2+1} d x \\
=x+\frac{1}{2} \log |x-1|+\frac{1}{4} \log \left|x^2+1\right|-\frac{1}{2} \tan ^{-1} x+C
\end{array}
$

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Question 124 Marks

Find the value of $\int \frac{\sin x}{\sqrt{1+\sin x}} d x$.

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Maths 4 Marks

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