Question

Integrate Sin4x/cos2x dx

Answer 1

Answer:

Step-by-step explanation:

I=∫sin4xcos2xdx  

=∫(1−cos(2x)2)21+cos(2x)2dx

=∫(1−cos(2x))22(1+cos(2x))dx

=∫12[(1+cos(2x))2−4cos(2x)1+cos(2x)]dx

=12∫[(1+cos(2x))−4cos(2x)1+cos(2x)]dx

=12[∫(1+cos(2x))dx−4∫(2cos2(x)−12cos2(x))dx]

=12[x+sin(2x)2−2∫(2−sec2(x))dx]

=12[x+sin(2x)2−4x+2tan(x)]+C

∴

I=sin(2x)4+tan(x)−32x+c

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