Maths MCQ

(c) : Given, $I_n=\int_0^{\pi / 4} \tan ^n \theta d \theta$
$
\begin{aligned}
& =\int_0^{\pi / 4} \tan ^{n-2} \theta\left(\tan ^2 \theta\right) d \theta=\int_0^{\pi / 4} \tan ^{n-2} \theta\left(\sec ^2 \theta-1\right) d \theta \\
& =\int_0^{\pi / 4} \tan ^{n-2} \theta \sec ^2 \theta d \theta-\int_0^{\pi / 4} \tan ^{n-2} \theta d \theta
\end{aligned}
$
Put $\tan \theta=t, \sec ^2 \theta d \theta=d t$, we get
$
\begin{aligned}
& \int_0^{\pi / 4} t^{n-2} d t-\int_0^{\pi / 4} \tan ^{n-2} \theta d \theta \\
& =\left[\frac{t^{n-1}}{n-1}\right]_0^{\pi / 4}-\int_0^{\pi / 4} \tan ^{n-2} \theta d \theta \\
& =\left[\frac{\tan ^{n-1} \theta}{n-1}\right]_0^{\pi / 4}-\int_0^{\pi / 4} \tan ^{n-2} \theta d \theta \\
& =\frac{1}{n-1}-\int_0^{\pi / 4} \tan ^{n-2} \theta d \theta \\
& \therefore I_n=\frac{1}{n-1}-I_{n-2} \\
& \therefore I_{12}=\frac{1}{12-1}-I_{12-2} \\
& \Rightarrow I_{12}+I_{10}=\frac{1}{11}
\end{aligned}
$

(b) : Let $I=\int_{\pi / 6}^{\pi / 3} \sec ^{2 / 3} x \operatorname{cosec}^{4 / 3} x d x$
$
\begin{aligned}
& =\int_{\pi / 6}^{\pi / 3} \frac{d x}{\sin ^{4 / 3} x \cos ^{2 / 3} x}=\int_{\pi / 6}^{\pi / 3} \frac{d x}{\left(\frac{\sin x}{\cos x}\right)^{4 / 3} \cos ^2 x} \\
& =\int_{\pi / 6}^{\pi / 3} \frac{\sec ^2 x}{(\tan x)^{4 / 3}} d x
\end{aligned}
$
Let $\tan x=t \Rightarrow \sec ^2 x d x=d t$
At $x=\frac{\pi}{6}, t=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}$ and at $x=\frac{\pi}{3}, t=\tan \frac{\pi}{3}=\sqrt{3}$
$
\therefore I=\int_{1 / \sqrt{3}}^{\sqrt{3}} \frac{1}{t^{4 / 3}} d t=\left[\frac{-3}{t^{1 / 3}}\right]_{1 / \sqrt{3}}^{\sqrt{3}}=\frac{-3}{(\sqrt{3})^{1 / 3}}+\frac{3}{\left(\frac{1}{\sqrt{3}}\right)^{1 / 3}}
$
$\begin{aligned} & =\frac{-3}{3^{1 / 6}}+\frac{3}{\frac{1}{3^{1 / 6}}}=-3^{1-\frac{1}{6}}+3^{1+\frac{1}{6}} \\ & =3^{7 / 6}-3^{5 / 6}\end{aligned}$

(c) : Let $I=\int_0^{\pi / 2} \frac{\sin x}{1+\cos ^2 x} d x$
Let $\cos x=t \Rightarrow-\sin x d x=d t$, when $x=0, t=1, x=\pi / 2$, $t=0$
$\Rightarrow I=-\int_1^0 \frac{d t}{1+t^2}=\left[-\tan ^{-1} t\right]_1^0=\tan ^{-1} 1=\frac{\pi}{4}$

(b) : Let $I=\int_0^{\pi / 4} \sec ^4 x d x$
$
I=\int_0^{\pi / 4} \sec ^2 x \cdot \sec ^2 x \cdot d x=\int_0^{\pi / 4}\left(1+\tan ^2 x\right) \sec ^2 x d x
$
Let $z=\tan x$, at $x=0, z=0$ and at $x=\pi / 4, z=1$
$
\begin{aligned}
& \frac{d z}{d x}=\sec ^2 x \Rightarrow d z=\sec ^2 x d x \\
& I=\int_0^1\left(1+z^2\right) d z=\left[z+\frac{z^3}{3}\right]_0^1=1+\frac{1}{3}=\frac{4}{3}
\end{aligned}
$

(b) : $I=\int_{-1}^1 \log \left(\frac{2-x}{2+x}\right) d x=\int_{-1}^1 f(x) d(x)$
$
f(x)=\log \left(\frac{2-x}{2+x}\right)
$
Now, $f(-x)=\log \left(\frac{2+x}{2-x}\right)=-\log \left(\frac{2-x}{2+x}\right)=-f(x)$
$\therefore f(x)$ is an odd function
$
\therefore \quad \int_{-1}^1 f(x) d x=0
$

$\begin{aligned} & \text {(b): } \int_0^{2 \pi}(\sin x+|\sin x|) d x \\
& =\int_0^{\pi / 2}(\sin x+\sin x) d x+\int_{\pi / 2}^\pi(\sin x+\sin x) d x \\
& +\int_\pi^{3 \pi / 2}(\sin x-\sin x) d x+\int_{3 \pi / 2}^{2 \pi}(\sin x-\sin x) d x \\
& =\int_0^\pi 2 \sin x d x=[-2 \cos x]_0^\pi \\ & =-2 \cos \pi-(-2 \cos 0)=2+2=4\end{aligned}$

$\begin{aligned} & \text {(d) : } \int_0^1 x(1-x)^n d x=\int_0^1(1-x)(1-(1-x))^n d x \\
& \left.\qquad \int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right] \\
& =\int_0^1(1-x) x^n d x=\int_0^1\left(x^n-x^{n+1}\right) d x \\
& =\left[\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}\right]_0^1=\frac{1}{n+1}-\frac{1}{n+2} \\
& =\frac{n+2-n-1}{(n+1)(n+2)}=\frac{1}{(n+1)(n+2)}\end{aligned}$

(a) : Let $I=\int_0^{\pi / 2} \log |\tan x+\cot x| d x$
$
\begin{aligned}
& =\int_0^{\pi / 2} \log \left|\frac{\sin ^2 x+\cos ^2 x}{\sin x \cos x}\right| d x \\
& =\int_0^{\pi / 2} \log \sin x d x-\int_0^{\pi / 2} \log \cos x d x=\pi \log 2
\end{aligned}
$

(d) : Let $I=\int_0^a \sqrt{\frac{a-x}{x}} d x$
Put $x=a \sin ^2 \theta \Rightarrow d x=2 a \sin \theta \cos \theta d \theta$
When $x=0, \theta=0$
When $x=a, \theta=\pi / 2$
$
\begin{aligned}
& \therefore I=\int_0^{\pi / 2} \sqrt{\frac{a\left(1-\sin ^2 \theta\right)}{a \sin ^2 \theta}} \cdot 2 a \sin \theta \cos \theta d \theta \\
& =2 a \int_0^{\pi / 2} \sqrt{\frac{\cos ^2 \theta}{\sin ^2 \theta}} \cdot \sin \theta \cdot \cos \theta d \theta=2 a \int_0^{\pi / 2} \cos ^2 \theta d \theta \\
& =2 a \int_0^{\pi / 2}\left(\frac{1+\cos 2 \theta}{2}\right) d \theta \\
& =2 a\left[\frac{\theta}{2}+\frac{\sin 2 \theta}{4}\right]_0^{\pi / 2}=2 a\left[\frac{\pi}{4}-0+0-0\right]=\frac{a \pi}{2}
\end{aligned}
$
Now, $\frac{a \pi}{2}=\frac{K}{2} \Rightarrow K=a \pi$

(c) : Let $I=\int_0^4 \frac{1}{1+\sqrt{x}} d x$
Put $1+\sqrt{x}=t \Rightarrow \frac{1}{2 \sqrt{x}} d x=d t$
$
\Rightarrow \quad d x=2 \sqrt{x} d t=2(t-1) d t
$
When $x=0, t=1$ and when $x=4, t=3$
$
\begin{aligned}
& \therefore \quad I=\int_1^3 \frac{2(t-1)}{t} \cdot d t=\int_1^3 2 \cdot d t-\int_1^3 \frac{2}{t} \cdot d t=2[t]_1^3-2[\log t]_1^3 \\
& =2 \times 2-2(\log 3-\log 1)=4-2 \log 3=4 \log e-\log 3^2 \\
& =\log e^4-\log 9=\log \left(\frac{e^4}{9}\right) \quad \therefore \quad I=\log \left(\frac{e^4}{9}\right)
\end{aligned}
$

(b): We have, $\int_{-3}^3\left(a x^5+b x^3+c x+k\right) d x$
$
\begin{aligned}
& =a \int_{-3}^3 x^5 d x+b \int_{-3}^3 x^3 d x+c \int_{-3}^3 x d x+k \int_{-3}^3 1 \cdot d x \\
& =0+0+0+k[x]_{-3}^3 \quad\left[\because \int_{-a}^a f(x) d x=0 \text { if } f(-x)=-f(x)\right]
\end{aligned}
$
$\therefore \quad$ The given integrand depends only on $k$.

(b) : Let $I=\int_0^1 x(1-x)^5 d x=-\int_0^1(1-x-1)(1-x)^5 d x$
$\begin{aligned} & =-\left[\int_0^1(1-x)^6 d x-\int_0^1(1-x)^5 d x\right]=\int_0^1(1-x)^5 d x-\int_0^1(1-x)^6 d x \\ & =\left[\frac{(1-x)^6}{-6}\right]_0^1-\left[\frac{(1-x)^7}{-7}\right]_0^1=\frac{1}{6}-\frac{1}{7}=\frac{1}{42}\end{aligned}$

$
\begin{aligned}
& \text {(b) : Let } I=\int_a^b \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a+b-x}} d x ...(i) \\
& \Rightarrow I=\int_a^b \frac{\sqrt{a+b-x}}{\sqrt{a+b-x}+\sqrt{x}} d x ..(ii) \\
& {\left[\because \int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right]}
\end{aligned}
$
Adding (i) and (ii), we get
$
\begin{aligned}
& 2 I=\int_a^b \frac{\sqrt{x}+\sqrt{a+b-x}}{\sqrt{x}+\sqrt{a+b-x}} d x=\int_a^b d x=[x]_a^b=b-a \\
& \Rightarrow I=\frac{b-a}{2}
\end{aligned}
$

 (b) : Let $I=\int_0^{\pi / 4} x \sec ^2 x d x$
Integrating by parts, we get
$
\begin{aligned}
I & =[x \tan x]_0^{\pi / 4}-\int_0^{\pi / 4} \tan x d x \\
& =[x \tan x]_0^{\pi / 4}+[\log |\cos x|]_0^{\pi / 4} \\
& =\frac{\pi}{4} \tan \frac{\pi}{4}-0+\log \left|\cos \frac{\pi}{4}\right|-\log |\cos 0| \\
& =\frac{\pi}{4} \times 1+\log \left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}+\log (1)-\log \sqrt{2}=\frac{\pi}{4}-\log (\sqrt{2})
\end{aligned}
$

$\begin{aligned} &  \text { (c) : Let } I=\int_0^K \frac{d x}{2+18 x^2}=\frac{1}{18} \int_0^K \frac{d x}{\frac{2}{18}+x^2} \\ & \Rightarrow I=\frac{1}{18} \int_0^K \frac{d x}{\frac{1}{9}+x^2}=\frac{1}{18} \int_0^K \frac{d x}{\left(\frac{1}{3}\right)^2+x^2} \\ & \Rightarrow I=\frac{1}{18}\left[\frac{1}{\left(\frac{1}{3}\right)} \tan ^{-1}\left(\frac{x}{\frac{1}{3}}\right)\right]_0^K=\frac{\pi}{24} \\ & \Rightarrow \tan ^{-1} 3 K=\frac{\pi}{4}=\tan ^{-1} 1 \Rightarrow 3 K=1 \Rightarrow K=\frac{1}{3} .\end{aligned}$

(d) : $\int_0^{\pi / 2} \log \sec x d x=\int_0^{\pi / 2} \log \left(\frac{1}{\cos x}\right) d x$
$=-\int_0^{\pi / 2} \log (\cos x) d x=-\frac{\pi}{2} \log (1 / 2)=\frac{\pi}{2} \log 2$

$\begin{aligned} & \text {(a) : } \int_0^3[x] d x=\int_0^1 0 d x+\int_1^2 1 d x+\int_2^3 2 d x \\ & =\left.x\right|_1 ^2+\left.2 x\right|_2 ^3=(2-1)+2(3-2)=1+2=3\end{aligned}$

(b) : We have, $\int_0^1 x \cdot \tan ^{-1} x d x$
$
\begin{aligned}
& =\left(\tan ^{-1} x \cdot \frac{x^2}{2}\right)_0^1-\int_0^1 \frac{1}{1+x^2} \cdot \frac{x^2}{2} d x \\
& =\left(\frac{\pi}{4} \cdot \frac{1}{2}-0\right)-\frac{1}{2} \int_0^1 \frac{1+x^2-1}{1+x^2} d x \\
& =\frac{\pi}{8}-\frac{1}{2} \int_0^1\left(1-\frac{1}{1+x^2}\right) d x=\frac{\pi}{8}-\frac{1}{2}\left[x-\tan ^{-1} x\right]_0^1 \\
& =\frac{\pi}{8}-\frac{1}{2}\left[\left(1-\frac{\pi}{4}\right)-(0)\right]=\frac{\pi}{8}-\frac{1}{2}+\frac{\pi}{8}=\frac{\pi}{4}-\frac{1}{2}
\end{aligned}
$

$
\begin{aligned}
& \text {(c) : Let } I=\int_0^{\pi / 2} \frac{\sqrt[n]{\sec x}}{\sqrt[n]{\sec x}+\sqrt[n]{\operatorname{cosec} x}} d x ...(i)\\
& =\int_0^{\pi / 2} \frac{\sqrt[n]{\sec \left(\frac{\pi}{2}-x\right)}}{\sqrt[n]{\sec \left(\frac{\pi}{2}-x\right)}+\sqrt[n]{\operatorname{cosec}\left(\frac{\pi}{2}-x\right)}} d x \\
& I=\int_0^{\pi / 2} \frac{\sqrt[n]{\operatorname{cosec} x}}{\sqrt[n]{\operatorname{cosec} x}+\sqrt[n]{\sec x}} d x...(ii)
\end{aligned}
$
Adding (i) and (ii), we get
$
2 I=\int_0^{\pi / 2} d x=[x]_0^{\pi / 2}=\frac{\pi}{2} \Rightarrow I=\frac{\pi}{4}
$

(d) : Let $I=\int_{-\pi / 2}^{\pi / 2} \log \left(\frac{2-\sin x}{2+\sin x}\right) d x$ Since, the given function is an odd function 
$\therefore \quad I=0$

$\begin{aligned} & \text {(a) : Let, } I=\int_0^{\pi / 2} \frac{\cos x}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^3} d x \\ & =\int_0^{\pi / 2} \frac{\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^3} d x=\int_0^{\pi / 2} \frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2} d x\end{aligned}$
Let $\cos \frac{x}{2}+\sin \frac{x}{2}=t \Rightarrow \frac{1}{2}\left(-\sin \frac{x}{2}+\cos \frac{x}{2}\right) d x=d t$
Also, $x=0 \Rightarrow t=1$ and $x=\frac{\pi}{2} \Rightarrow t=\sqrt{2}$
$
\therefore \quad I=\int_1^{\sqrt{2}} \frac{2 d t}{t^2}=2\left[-\frac{1}{t}\right]_1^{\sqrt{2}}=2\left[-\frac{1}{\sqrt{2}}+1\right]=2-\sqrt{2}
$

 (c) : Let $\tan ^{-1} x=\theta \Rightarrow x=\tan \theta$
$
\Rightarrow d x=\sec ^2 \theta d \theta
$
Now, $x=0 \Rightarrow \theta=0$ and $x=1 \Rightarrow \theta=\frac{\pi}{4}$
$
\begin{aligned}
& \therefore \quad \int_0^1 \frac{x \tan ^{-1} x}{\left(1+x^2\right)^{3 / 2}} d x=\int_0^{\pi / 4} \frac{\theta \tan \theta}{\sec ^3 \theta} \sec ^2 \theta d \theta \\
& =\int_0^{\pi / 4} \theta \sin \theta d \theta=[-\theta \cos \theta]_0^{\pi / 4}-\int_0^{\pi / 4}(-\cos \theta) d \theta \\
&
\end{aligned}
$
$=[-\theta \cos \theta]_0^{\pi / 4}+[\sin \theta]_0^{\pi / 4}=\frac{4-\pi}{4 \sqrt{2}}$