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}.\frac{sinx}{\cancel{cosx}}dx`

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

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

`\\ \sf\longmapsto {\displaystyle{\int}}(1-cos2x)/(2)dx`

`\\ \sf\longmapsto (1)/(2){\displaystyle{\int}}1-cos2x`

`\\ \sf\longmapsto (1)/(2){\displaystyle{\int}}dx-(1)/(2){\displaystyle{\int}}cos2x`

`\\ \sf\longmapsto (1)/(2)x-(1)/(2){\displaystyle{\int}}cos2xdx`

Here Taking

• u=2x

Then

• du=2dx

Now put values

`\\ \sf\longmapsto (1)/(2)du=dx`

Then

`\\ \sf\longmapsto (1)/(2)x-(1)/(2){\displaystyle{\int}}cos2xdx`

`\\ \sf\longmapsto (1)/(2)x-(1)/(2)* (1)/(2){\displaystyle{\int}}cosudu`

`\\ \sf\longmapsto (1)/(2)x-(1)/(4){\displaystyle{\int}}cosudu`

`\\ \sf\longmapsto (1)/(2)x-(1)/(4)sin u`

`\\ \sf\longmapsto\underline{\boxed{\bf{(1)/(2)x-(1)/(4)sin 2x+C}}}`