Maths MCQ

$=\int \frac{\sec ^2 x d x}{\sqrt{\tan ^2 x-1}} \quad \ldots$. [Dividing by $\cos ^2 x$ ]

Put $\tan x=t$.

Hint: $\frac{d}{d x}(x x)=x^x(1+\log x)$

Hint : $\int \frac{1}{\cos x-\cos ^2 x} d x$

$=\int \frac{1}{\cos x(1-\cos x)} d x$

$=\int \frac{(1-\cos x)+\cos x}{\cos x(1-\cos x)} d x$

$=\int\left[\sec x+\frac{1}{2} \operatorname{cosec}^2\left(\frac{x}{2}\right)\right] d x$

$=\log |\sec x+\tan x|+\frac{1}{2} \frac{\left(-\cot \frac{x}{2}\right)}{1 / 2}+c$

$=\log |\sec x+\tan x|-\cot \left(\frac{x}{2}\right)+c$

Hint: $\int \tan ^3 x \cdot \sec ^3 x d x$

$=\int \sec ^2 x \cdot \tan ^2 x \cdot \sec x \tan x d x$

$=\int \sec ^2 x\left(\sec ^2 x-1\right) \sec x \tan x d x$

Put sec x = t.

Hint : $\int \frac{x^4}{x+x^5} d x=\int \frac{\left(x^4+1\right)-1}{x\left(x^4+1\right)} d x$

$=\int\left(\frac{1}{x}-\frac{1}{x+x^5}\right) d x=\log x-f(x)+c$

$
\begin{aligned}
& \text { (b) : } I=\int \frac{\sin x+\sin ^3 x}{\cos 2 x} d x \\
& =\int \frac{\sin x\left(1+\sin ^2 x\right)}{2 \cos ^2 x-1} d x \\
& =\int \frac{\sin x\left(1+1-\cos ^2 x\right)}{2 \cos ^2 x-1} d x \\
& =\int \frac{\left(2-\cos ^2 x\right) \sin x}{2 \cos ^2 x-1} d x
\end{aligned}
$

Let $\cos x=t$, then $\sin x d x=-d t$
$
\begin{aligned}
& \therefore \quad I=-\int \frac{2-t^2}{2 t^2-1} d t \\
& =\int \frac{t^2-2}{2 t^2-1} d t=\frac{1}{2} \int \frac{2\left(t^2-2\right)}{2 t^2-1} d t \\
& =\frac{1}{2} \int \frac{2 t^2-1-3}{2 t^2-1} d t=\frac{1}{2} \int d t-\frac{3}{2} \int \frac{1}{2 t^2-1} d t \\
& =\frac{1}{2} t-\frac{3}{4} \int \frac{1}{t^2-\left(\frac{1}{\sqrt{2}}\right)^2} d t \\
& =\frac{1}{2} t-\frac{3}{4} \cdot \frac{1}{2 \times \frac{1}{\sqrt{2}}} \log \left|\frac{t-\frac{1}{\sqrt{2}}}{t+\frac{1}{\sqrt{2}}}\right|+c \\
& =\frac{1}{2} \cos x-\frac{3}{4 \sqrt{2}} \log \left|\frac{\sqrt{2} \cos x-1}{\sqrt{2} \cos x+1}\right|+c \\
& \therefore \quad P=\frac{1}{2} \text { and } Q=\frac{-3}{4 \sqrt{2}}
\end{aligned}
$

$\begin{aligned} & \text {(b) : Let } I=\int \frac{d x}{\sin (x-a) \sin x} \\ & =\int \frac{\sin a}{\sin a \cdot \sin x \sin (x-a)} d x \\ & =\frac{-1}{\sin a} \int \frac{\sin (x-a-x)}{\sin x \sin (x-a)} d x \\ & =\frac{-1}{\sin a} \int \frac{\sin (x-a) \cos x-\cos (x-a) \sin x}{\sin x \sin (x-a)} d x \\ & =\frac{-1}{\sin a} \int[\cot x-\cot (x-a)] d x\end{aligned}$
$\begin{aligned} & =\frac{-1}{\sin a}[\ln (\sin x)-\ln \sin (x-a)]+C \\ & =\operatorname{cosec} a\left[\ln \sin (x-a)-\ln \left(\frac{1}{\operatorname{cosec} x}\right)\right]+C \\ & =\operatorname{cosec} a[\ln (\sin (x-a) \cdot \operatorname{cosec} x]+C\end{aligned}$

(b) : Let $I=\int \frac{1}{\cot ^2 x-1} d x$
$
\begin{aligned}
& =\int \frac{\sin ^2 x}{\cos ^2 x-\sin ^2 x} d x=\int \frac{\sin ^2 x}{\cos 2 x} d x=\int \frac{1-\cos 2 x}{2 \cos 2 x} d x \\
& =\frac{1}{2} \int\left(\frac{1}{\cos 2 x}-1\right) d x=\frac{1}{2} \int(\sec 2 x-1) d x \\
& =\frac{1}{2} \frac{\log |\sec 2 x+\tan 2 x|}{2}-\frac{x}{2}+c \\
& =\frac{1}{4} \log |\sec 2 x+\tan 2 x|-\frac{x}{2}+c
\end{aligned}
$
(where ' $c$ ' is constant of integration)
On comparing with equation, we get
$
A=4 \text { and } B=2 \quad \therefore A+B=4+2=6
$

$\begin{aligned} & \text {(d) : Let } I=\int \frac{d x}{\sin (x-a) \sin (x-b)} \\ & =\frac{1}{\sin (b-a)} \int \frac{\sin (b-a)}{\sin (x-a) \sin (x-b)} d x\end{aligned}$
$\begin{aligned} & =\frac{1}{\sin (b-a)} \int \frac{\sin ((x-a)-(x-b))}{\sin (x-a) \sin (x-b)} d x \\ & =\frac{1}{\sin (b-a)} \int \frac{\sin (x-a) \cos (x-b)-\cos (x-a) \sin (x-b)}{\sin (x-a) \sin (x-b)} d x \\ & =\frac{1}{\sin (b-a)} \int(\cot (x-b)-\cot (x-a)) d x \\ & =\frac{1}{\sin (b-a)}[\log |\sin (x-b)|-\log |\sin (x-a)|]+c \\ & =\frac{1}{\sin (b-a)} \log \left|\frac{\sin (x-b)}{\sin (x-a)}\right|+c\end{aligned}$

$\begin{aligned} & \text {(a) : Let } I=\int \frac{x^2}{\sqrt{1-x}} d x \\ & \text { Let } 1-x=t^2 \\ & \Rightarrow \quad x=1-t^2 \text { and } d x=-2 t d t \\ & \therefore I=-2 \int \frac{t\left(1-t^2\right)^2}{t} d t=-2 \int\left(1-2 t^2+t^4\right) d t \\ & =-2\left(t-\frac{2 t^3}{3}+\frac{t^5}{5}\right)+c=-2\left[\frac{15 t-10 t^3+3 t^5}{15}\right]+C \\ & =-2 \sqrt{1-x}\left[\frac{15-10(1-x)+3(1-x)^2}{15}\right]+C \\ & =-\frac{2 \sqrt{1-x}}{15}\left(3 x^2+4 x+8\right) \quad \therefore \quad p=\frac{-2}{15}[\text { using (i) }]\end{aligned}$

$
\begin{aligned}
& \text {(d) : Let } I=\int \frac{\cos \theta}{5+7 \sin \theta-2 \cos ^2 \theta} d \theta \\
& =\int \frac{\cos \theta}{5+7 \sin \theta-2\left(1-\sin ^2 \theta\right)} d \theta \\
& =\int \frac{\cos \theta}{2 \sin ^2 \theta+7 \sin \theta+3} d \theta=\int \frac{\cos \theta}{(2 \sin \theta+1)(\sin \theta+3)} d \theta
\end{aligned}
$
Let $\sin \theta=t \Rightarrow \cos \theta d \theta=d t$
$
\begin{aligned}
\therefore I & =\int \frac{d t}{(2 t+1)(t+3)}=\frac{1}{5} \int\left(\frac{2}{2 t+1}-\frac{1}{t+3}\right) d t \\
& =\frac{1}{5}[\log |2 t+1|-\log |t+3|]=\frac{1}{5} \log \left|\frac{2 t+1}{t+3}\right| \\
& =\frac{1}{5} \log \left|\frac{2 \sin \theta+1}{\sin \theta+3}\right| \\
\therefore A & =\frac{1}{5} \text { and } f(\theta)=\frac{2 \sin \theta+1}{\sin \theta+3} \quad \therefore \frac{f(\theta)}{A}=\frac{5(2 \sin \theta+1)}{\sin \theta+3}
\end{aligned}
$

$
\begin{aligned}
& \text {(b) : Let } I=\int \frac{\sin 2 x\left(1-\frac{3}{2} \cos x\right)}{e^{\sin ^2 x+\cos ^3 x}} d x \\
& =\int \frac{2 \sin x \cos x-3 \cos ^2 x \sin x}{e^{\sin ^2 x+\cos ^3 x}} d x[\because \sin 2 x=2 \sin x \cos x]
\end{aligned}
$
Let $\sin ^2 x+\cos ^3 x=t$
$
\begin{aligned}
& \Rightarrow\left(2 \sin x \cos x-3 \cos ^2 x \sin x\right) d x=d t \\
& \Rightarrow I=\int \frac{d t}{e^t}=\int e^{-t} d t=-e^t+c \\
& =-e^{-\left(\sin ^2 x+\cos ^3 x\right)}+c \\
&
\end{aligned}
$

(b) : $I=\int \frac{x^3 d x}{\sqrt{1+x^2}}=\int \frac{x^2 \cdot x}{\sqrt{1+x^2}} d x$
Let $\left(1+x^2\right)=t^2 \Rightarrow 2 x d x=2 t d t \Rightarrow x d x=t d t$
$
\begin{aligned}
& \therefore I=\int \frac{t\left(t^2-1\right)}{t} d t=\int\left(t^2-1\right) d t \\
& =\frac{t^3}{3}-t+c=\frac{1}{3}\left(1+x^2\right)^{3 / 2}-\sqrt{1+x^2}+c \\
& =\frac{1}{3}\left(1+x^2\right) \sqrt{1+x^2}-\sqrt{1+x^2}+c \\
& \therefore \quad a=\frac{1}{3} \text { and } b=-1 \Rightarrow 3 a b=\frac{-3}{3}=-1
\end{aligned}
$

$
\begin{aligned}
& \text {(a) : Let } I=\int \frac{\sin x+\sin ^3 x}{\cos 2 x} d x \\
& =\int \frac{\sin x+\left(1+\sin ^2 x\right)}{2 \cos ^2 x-1} d x=\int \frac{\sin x\left(1+1-\cos ^2 x\right)}{2 \cos ^2 x-1} d x \\
& =\int \frac{\sin x\left(2-\cos ^2 x\right)}{2 \cos ^2 x-1} d x
\end{aligned}
$
Let $\cos x=t \Rightarrow-\sin x d x=d t$
$
\begin{aligned}
& \therefore \quad I=\int \frac{-\left(2-t^2\right)}{2 t^2-1} d t=\frac{1}{2} \int \frac{2\left(t^2-2\right)}{2 t^2-1} d t \\
& =\frac{1}{2} \int \frac{2 t^2-1-3}{2 t^2-1} d t=\frac{1}{2} \int\left(1-\frac{3}{2 t^2-1}\right) \\
& =\frac{1}{2} \int 1 d t-\frac{3}{4} \int \frac{1}{t^2-\left(\frac{1}{\sqrt{2}}\right)^2} d t \\
& =\frac{1}{2} t-\frac{3}{4 \sqrt{2}} \log \left|\frac{\sqrt{2} t-1}{\sqrt{2} t+1}\right|+c \\
& =\frac{1}{2} \cos x-\frac{3}{4 \sqrt{2}} \log \left|\frac{\sqrt{2} \cos x-1}{\sqrt{2} \cos x+1}\right|+c \\
& \therefore A=\frac{1}{2}, B=-\frac{3}{4 \sqrt{2}}, f(x)=\frac{\sqrt{2} \cos x-1}{\sqrt{2} \cos x+1}
\end{aligned}
$

(b) : $I=\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x$
Let $\sqrt{x}=\sin t \Rightarrow x=\sin ^2 t$
$\Rightarrow d x=2 \sin t \cos t d t$
$\therefore \quad I=2 \int \sqrt{\frac{1-\sin t}{1+\sin t}} \cdot \sin t \cos t d t$
$=2 \int \sqrt{\frac{(1-\sin t)(1-\sin t)}{1-\sin ^2 t}} \sin t \cos t d t$
$=2 \int \frac{1-\sin t}{\cos t} \times \sin t \cos t d t=2 \int\left(\sin t-\sin ^2 t\right) d t$
$=2 \int\left(\sin t-\left(\frac{1-\cos 2 t}{2}\right)\right) d t$
$=2\left[-\cos t-\frac{1}{2} t+\frac{\sin 2 t}{4}\right]+C$
$=-2 \cos t-t+\sin t \cos t+C$
$=-2 \sqrt{1-x}-\sin ^{-1} \sqrt{x}+\sqrt{x} \sqrt{1-x}+C$
$=-2 \sqrt{1-x}+\cos ^{-1} \sqrt{x}+\sqrt{x(1-x)}+C_1$
$\left[\because \sin ^{-1} x+\cos ^{-1}=\frac{\pi}{2}\right]$

(c) : Let $I=\int \frac{\sin 4 x}{\sin x} d x$
We know $\sin (A+B)=\sin A \cos B+\cos A \sin B$
So, $\sin (2 x+2 x)=\sin 2 x \cos 2 x+\cos 2 x \sin 2 x$
$
=2 \sin 2 x \cos 2 x=4 \sin x \cos x \cos 2 x
$
$
\begin{aligned}
I=\int & \frac{4 \sin x \cos x \cos 2 x}{\sin x} d x=\int 4 \cos x \cos 2 x d x \\
=\int 2(\cos x+\cos 3 x) d x & =2 \sin x+\frac{2}{3} \sin 3 x+C
\end{aligned}
$

(a) : Let $I=\int \frac{d x}{2+\cos x}$
$=\int \frac{d x}{2+\frac{1-\tan ^2 x / 2}{1+\tan ^2 x / 2}}=\int \frac{\sec ^2 x / 2}{3+\tan ^2 x / 2} d x$
Let $z=\tan \frac{x}{2}$
$\frac{d z}{d x}=\frac{1}{2} \sec ^2 x / 2 ; \sec ^2 \frac{x}{2} d x=2 d z$
So, $I=\int \frac{2 d z}{(\sqrt{3})^2+z^2}=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{z}{\sqrt{3}}\right)+C$
Now substitute $z=\tan \frac{x}{2}$
$I=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{\tan \left(\frac{x}{2}\right)}{\sqrt{3}}\right)+C$

(c) : Let $I=\int \frac{x^3+x}{x^4-9} d x$
$=\int \frac{x^3}{x^4-9} d x+\int \frac{x}{x^4-9} d x=I_1+I_2+C$ (say), where
$I_1=\int \frac{x^3}{x^4-9} d x$ and $I_2=\int \frac{x}{x^4-9} d x$
Putting $x^4-9=t$ in $I_1$
$\Rightarrow 4 x^3 d x=d t$, we get
$I_1=\frac{1}{4} \int \frac{1}{t} d t=\frac{1}{4} \log |t|=\frac{1}{4} \log \left|x^4-9\right|$
Now, $I_2=\int \frac{x}{x^4-9} d x=\int \frac{x}{\left(x^2\right)^2-3^2} d x$
Putting $x^2=t \Rightarrow 2 x d x=d t$, we get
$I_2=\frac{1}{2} \int \frac{d t}{t^2-3^2}=\frac{1}{2} \cdot \frac{1}{2 \times 3} \log \left|\frac{t-3}{t+3}\right|=\frac{1}{12} \log \left|\frac{x^2-3}{x^2+3}\right|
$
Hence, $I=\frac{1}{4} \log \left|x^4-9\right|+\frac{1}{12} \log \left|\frac{x^2-3}{x^2+3}\right|+C$