How to evaluate this indefinite integral #∫[10sin(x)]/[1+cos^2(x)]dx#?

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Answer 1 · 2017-12-02T07:15:16

The answer is #=-10arctan(cosx)+C#

Perform this integral by substitution

Let #u=cosx#, #=>#, #du=-sinxdx#

Therefore,

#int(10sinxdx)/(1+cos^2x)=-10int(du)/(1+u^2)#

Let #u=tantheta#

#du=sec^2thetad theta#

#1+tan^2theta=sec^2theta#

So,

#int(10sinxdx)/(1+cos^2x)=-10int(sec^2thetad theta)/(1+tan^2theta)#

#=-10int(sec^2thetad theta)/(sec^2theta)#

#=-10intd theta#

#=-10theta#

#=-10rctanu#

#=-10arctan(cosx)+C#