Question

Fill in the blanks.
$\left(i\right)\int \frac{x+3}{{\left(x+4\right)}^2}{e}^{x}dx=P$.
$\left(ii\right)$ The value of $\underset{-\pi }{\overset{\pi }{\int }}si{n}^{3}xco{s}^{2}xdx$ is $Q$.
$\left(iii\right)\underset{0}{\overset{\frac{\pi }{2}}{\int }}cosx{e}^{sinx}dx$ is equal to $R$ .
$\left(iv\right)\int \frac{sinx}{3+4co{s}^{2}x}dx=S$.

	P	Q	R	S
(a)	$\left(\frac{e^x}{x+4}\right)+C$	$0$	$e-1$	$\frac{-1}{2\sqrt{3}}ta{n}^{-1}\left(\frac{2cosx}{\sqrt{3}}\right)+C$
(b)	$\frac{e^x}{\left(x+4\right)}+C$	$1$	$e-1$	$\frac{1}{\sqrt{3}}ta{n}^{-1}\left(\frac{2cosx}{\sqrt{3}}\right)+C$
(c)	$\frac{e^x}{{\left(x+4\right)}^2}+C$	$0$	$e+1$	$\frac{-1}{2\sqrt{3}}ta{n}^{-1}\left(\frac{2sinx}{\sqrt{3}}\right)+C$
(d)	$\frac{e^x+1}{\left(x+4\right)}+C$	$1$	$e+1$	$\frac{1}{\sqrt{3}}ta{n}^{-1}\left(\frac{2sinx}{\sqrt{3}}\right)+C$

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (A)

$\left(i\right)$ Let $I=\int \frac{x+3}{{\left(x+4\right)}^2}{e}^{x}dx$
$=\int e^x\left(\frac{1}{\left(x+4\right)}-\frac{1}{{\left(x+4\right)}^2}\right)dx=e^x\left(\frac{1}{x+4}\right)+C$
$[$Using $\int e^x\left\{f\left(x\right)+f^{\prime }\left(x\right)\right\}dx=e^{x}f\left(x\right)+C]$
$\left(ii\right)$ Let $f\left(x\right)=\underset{-\pi }{\overset{\pi }{\int }}si{n}^{3}xco{s}^{2}xdx$
Now, $f\left(-x\right)=\underset{-\pi }{\overset{\pi }{\int }}si{n}^3\left(-x\right)co{s}^2\left(-x\right)dx$
$=\underset{-\pi }{\overset{\pi }{\int }}-si{n}^{3}xco{s}^{2}xdx=-f\left(x\right)$
since, $f\left(x\right)$ is an odd function. ,
$\therefore \underset{-\pi }{\overset{\pi }{\int }}si{n}^{3}xco{s}^{2}xdx=0$
$\left(iii\right)$ Let $I=\underset{0}{\overset{\pi /2}{\int }}cosx{e}^{sinx}dx$
Substitute $sinx=t\Rightarrow cosxdx=dt$
When $x\to 0$
$\Rightarrow t\to 0$ and
when $x\to \pi /2\Rightarrow t\to 1$
$\therefore I=\underset{0}{\overset{1}{\int }}{e}^{t}dt={\left[e^t\right]}_0^1$
$=e^1-e^0=e-1$
$\left(iv\right)$ Let $I=\int \frac{sinx}{3+4co{s}^{2}x}dx$
Substitute $cosx=t$
$\Rightarrow -sinxdx=dt$
$\therefore \int \frac{dt}{3+4t^2}=-\frac{1}{4}\int \frac{dt}{{\left(\frac{\sqrt{3}}{2}\right)}^2+t^2}$
$-\frac{1}{4}\cdot \frac{2}{\sqrt{3}}ta{n}^{-1}\frac{2t}{\sqrt{3}}+C$
$=-\frac{1}{2\sqrt{3}}ta{n}^{-1}\left(\frac{2cosx}{\sqrt{3}}\right)+C$