Question

Evaluate these questions in the image ..

Answer 1

There are many.  It requires a lot steps to write.  I will give directions to solve that.  Follow them to get the answers.

2. 
$f(x)dx=sec^{\frac{4}{3}}x\ cosec^{\frac{8}{3}}x\ dx\\\\let\ secx=y,\ \ dx=\frac{dy}{y\sqrt{y^2-1}},\ cosecx=\frac{y}{\sqrt{y^2-1}}\\\\f(x)dx=\frac{y^3\ dy}{(y^2-1)^{\frac{11}{6}}}\\\\let\ y^2-1=z^2,\ dy=\frac{z\ dz}{\sqrt{z^2+1}}\\\\f(x)dx=(z^{-\frac{2}{3}}+z^{-\frac{8}{3}}dz\\\\integrate\ and\ substitute\\\\I=\frac{3}{5}\frac{5sec^2-6}{tan^{\frac{5}{3}}x}$

you could verify it by differentiating the expression..  you could also assume y = tan x  in the beginning.
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3.
use integration by parts method.      v ' =1,  so  v = x
$u=sin^{-1}\{ \frac{x}{x+a} \},\ \ du=\frac{a\ dx}{(x+a)\sqrt{a^2+2ax}}=\frac{2a\ dw}{a^2+w^2},\\\\where\ w^2=a^2+2ax,\ dx=w\ dw/a,\ x=(w^2-a^2)/2a\\\\I=x\ sin^{-1}\{ \frac{x}{x+a} \}- \int\limits^{}_{} {[ 1-\frac{2a^2}{a^2+w^2} ]} \, dw\\\\= x\ sin^{-1}\{ \frac{x}{x+a} \}-\sqrt{a(a+2x)}+2a\ tan^{-1}\{ \sqrt(1+\frac{2x}{a} \}+K$

4. 
$f(x)dx= \frac{\sqrt2\ Sin(x-a)\ dx}{\sqrt{2Sin(x-a)\ Sin(x+a)}}\\\\=\frac{\sqrt2\ (sinx\ cosa-cosx\ sina)}{\sqrt{cos2a-cos2x}}dx\\\\=\frac{\sqrt2\ cosa\ sinx\ dx}{\sqrt{cos2a+1-2cos^2x}}-\frac{\sqrt2\ sina\ cosx}{\sqrt{2sin^2x-(1-cos2a)}}\\\\\frac{sinx\ dx}{\sqrt{1-(\frac{cosx}{cosa})^2}}-\frac{cosx\ dx}{\sqrt{(\frac{sinx}{sina})^2-1}}\\\\integrating,\ I=-cosa\ sin^{-1}(\frac{cosx}{cosa})-sina\ cosh^{-1}(\frac{sinx}{sina})\\\\.$

Note that the  cosh  hyperbolic inverse function can also be expressed as a logarithm too.