Question

Topic :- Calculus (integral)

Solve the integration:-
∫ (sin x. cos x. tan x) dx​

Answer 1

$\\ \sf\longmapsto {\displaystyle{\int}}(sinx.cosx.tanx) dx$

$\\ \sf\longmapsto {\displaystyle{\int}}sinx.\cancel{cosx}.\dfrac{sinx}{\cancel{cosx}}dx$

$\\ \sf\longmapsto {\displaystyle{\int}}sinx.sinx \:dx$

$\\ \sf\longmapsto {\displaystyle{\int}}sin^2x\:dx$

$\\ \sf\longmapsto {\displaystyle{\int}}\dfrac{1-cos2x}{2}dx$

$\\ \sf\longmapsto \dfrac{1}{2}{\displaystyle{\int}}1-cos2x$

$\\ \sf\longmapsto \dfrac{1}{2}{\displaystyle{\int}}dx-\dfrac{1}{2}{\displaystyle{\int}}cos2x$

$\\ \sf\longmapsto \dfrac{1}{2}x-\dfrac{1}{2}{\displaystyle{\int}}cos2xdx$

Here Taking

• u=2x

Then

• du=2dx

Now put values

$\\ \sf\longmapsto \dfrac{1}{2}du=dx$

Then

$\\ \sf\longmapsto \dfrac{1}{2}x-\dfrac{1}{2}{\displaystyle{\int}}cos2xdx$

$\\ \sf\longmapsto \dfrac{1}{2}x-\dfrac{1}{2}\times \dfrac{1}{2}{\displaystyle{\int}}cosudu$

$\\ \sf\longmapsto \dfrac{1}{2}x-\dfrac{1}{4}{\displaystyle{\int}}cosudu$

$\\ \sf\longmapsto \dfrac{1}{2}x-\dfrac{1}{4}sin u$

$\\ \sf\longmapsto\underline{\boxed{\bf{\dfrac{1}{2}x-\dfrac{1}{4}sin 2x+C}}}$