Question 1

Topic :- Calculus (integral)

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

Question 2

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

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

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

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

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

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

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

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

Here Taking

Then

Now put values

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

Then

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

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

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

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

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

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