Question

Column -I and Column -II contains four entries each. Entry of column -I are to be matched one or more than one entries of column -II.

Answer 1

$A=\int \frac{cosx-sinx}{cosx+sinx}dx$
Substituting $t=sinx+cosx\Rightarrow dt=(cosx-sinx)dx$
$A=\int \frac{1}{t}dt=\log t+c=\log (cosx+sinx)+ccosx+sinx=\sqrt{2}(\frac{1}{\sqrt{2}}cosx+\frac{1}{\sqrt{2}}sinx)=\sqrt{2}\left(cos(x-\frac{\pi }{4})\right)$
$A=\log \left(\sqrt{2}\left(cos(x-\frac{\pi }{4})\right)\right)+c=\log \left(cos(x-\frac{\pi }{4})\right)+\log \sqrt{2}+c=\log \left(cos(x-\frac{\pi }{4})\right)+c$
$B=\int \frac{cos2x}{sin2x+1}dx$
Substituting $t=2x\Rightarrow dt=2dx$, we get
$B=\frac{1}{2}\int \frac{cost}{sint+1}dt$
Again substituting $sint+1=u\Rightarrow costdt=du$
$B=\frac{1}{2}\int \frac{1}{u}du=\frac{\log u}{2}+c=\frac{1}{2}\log (sint+1)+c=\log (sin2x+1)+c=\log (sinx+cosx)+c=\log \left(cos(x-\frac{\pi }{4})\right)+c$
$C=\int \frac{1-sin2x}{cos2x}dx=\int (1-sin2x)sec2xdx$
Substituting $t=2x\Rightarrow dt=2dx$
$C=\frac{1}{2}\int (1-sint)sectdt=\int (sect+tant)dt=-\frac{1}{2}\int tantdt+\frac{1}{2}\int sectdt=\frac{1}{2}\int \frac{sint}{cost}dt+\frac{1}{2}\int sectdt$
Again substituting $cost=u\Rightarrow -sintdt=du$
$C=-\frac{1}{2}\int -\frac{1}{u}du+\frac{1}{2}\int sectdt=\frac{1}{2}\int \frac{1}{u}du+\frac{1}{2}\int sectdt=\frac{\log u}{2}+\frac{1}{2}\log (tant+sect)+c=\frac{1}{2}\log \left(cos2x\right)+\frac{1}{2}\log (tant+sect)+c=\frac{1}{2}\log (sin2x+1)+c=\log (cosx+sinx)+c=\log \left(cos(x-\frac{\pi }{4})\right)+c$
$D=\int \left(\frac{si{n}^{-1}\sqrt{x}}{\sqrt{x-x^2}}\right)dx$
Substituting $t=si{n}^{-1}\sqrt{x}\Rightarrow dt=\frac{1}{2\sqrt{-x(x-1)}}dx$
$D=2\int tdt=t^2+c={\left(si{n}^{-1}\sqrt{x}\right)}^2+c={(\pi -co{s}^{-1}\sqrt{x})}^2+c={\pi }^2+{\left(co{s}^{-1}\sqrt{x}\right)}^2-2\pi co{s}^{-1}\sqrt{x}+c={\left(co{s}^{-1}\sqrt{x}\right)}^2-2\pi co{s}^{-1}\sqrt{x}+c$