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 $\int \limits^{\pi}_{-\pi} sin^{3} \,x \,cos^{2} x\, dx$ is $Q$.
$\left(iii\right) \int \limits_{0}^{\frac{\pi}{2}} cos\, x\, e^{ sin \,x} dx$ is equal to $R$ .
$\left(iv\right) \int\frac{sin \,x}{3+ 4\, cos^{2}x } dx = S$.

	P$ \quad$	Q $ \quad$	R $ \quad$	S$ \quad$
(a) $ \quad$	$\left(\frac{e^{x}}{x+4}\right) + C $	$0$	$e-1$	$\frac{-1}{2\sqrt{3}} tan^{-1} \left(\frac{2\,cos x}{\sqrt{3}}\right) + C$
(b)	$\frac{e^{x }}{\left(x+4\right)} +C$	$1$	$e-1$	$\frac{1}{\sqrt{3}} tan^{-1} \left(\frac{2\,cos\, x}{\sqrt{3}}\right) + C$
(c)	$\frac{e^{x }}{\left(x+4\right)^{2}} + C$	$0$	$e+1$	$\frac{-1}{2\sqrt{3}} tan^{-1} \left(\frac{2\,sin\, x}{\sqrt{3}}\right) + C$
(d)	$\frac{e^{x }+1}{\left(x+4\right)} +C$	$1$	$e+1$	$\frac{1}{\sqrt{3}} tan^{-1} \left(\frac{2\,sin\, x}{\sqrt{3}}\right) + C$

• 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'\left(x\right)\right\} dx = e^{x} f\left(x\right) + C] $
$\left(ii\right)$ Let $f\left(x\right) = \int\limits_{-\pi}^{\pi} sin^{3} x \,cos^{2} x dx$
Now, $f\left(-x\right) = \int\limits_{-\pi }^{\pi } sin^{3} \left(-x\right) cos^{2} \left(-x\right) \,dx $
$= \int\limits_{-\pi }^{\pi }-sin^{3} x \,cos^{2}x\,dx = -f\left(x\right)$
since, $f\left(x\right)$ is an odd function. ,
$\therefore \int\limits_{-\pi }^{\pi }sin^{3} x\, cos^{2}xdx = 0$
$ \left(iii\right)$ Let $I= \int\limits_{0}^{\pi /2} cos \,x\, e^{ sin \,x} dx$
Substitute $sin\,x = t \Rightarrow cos\,x\, dx = dt$
When $x \rightarrow 0 $
$\Rightarrow t \rightarrow 0$ and
when $x\rightarrow\pi/2 \Rightarrow t \rightarrow 1 $
$ \therefore I = \int\limits_{0}^{1} e^{t} dt = \left[ e^{t}\right]_{0}^{1} $
$=e^{1} -e^{0} = e-1$
$\left(iv\right)$ Let $I= \int \frac{sin\,x}{3+4 \,cos^{2} x } dx $
Substitute $cos\, x =t $
$\Rightarrow -sin\,x \,dx = dt$
$ \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}} tan^{-1} \frac{2t}{\sqrt{3}} + C $
$ = -\frac{1}{2\sqrt{3}} tan^{-1} \left(\frac{2cos x}{\sqrt{3}}\right) +C $