Question

Integrate Tan^2x * sin^2x * dx

pls help to me find out​

Answer 1

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Answer:

∫sin(2x)tan(2x)dx=12∫sin2(2x)1−sin2(2x)cos(2x)d(2x)

Make the substitution sin(2x)=t⟹cos(2x)d(2x)=dt. Thus,

∫sin2(2x)1−sin2(2x)cos(2x)d(2x)=t21−t2dt

=∫11−t2dt−∫dt=12ln∣∣∣1+t1−t∣∣∣−t+C

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Answer 2

Answer:

The answer is $\int\ tanx-3/2*x+sin2x/4+C$

Step-by-step explanation:

• To integrate the following equation we will follow the following steps:

                  =$\int\tan^2sin^2 {x} \, dx$=$\int\sin^4 {x}/cos^2*dx$

                  = $\int\sin^2x(1-cos^2x/cos^2x). \, dx$

                   =$\int\ (sec^2x-1-1-cos2x/2). \, dx$

                   =$\int\ tanx-3/2*x+sin2x/4+C$

Therefore, the answer is $\int\ tanx-3/2*x+sin2x/4+C$

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