Question

integration dx/3 sin2 x+4 cos2x​

Answer 1

Answer:

I=16arctan(12cos(x))−13arctan(cos(x))+C

Step-by-step explanation:

Let

I=∫sin(x)(4+cos2(x))(2−sin2(x))dx

First rewrite the second bracket in the denominator to convert the sin2(x) term into cos2(x):

I=∫sin(x)(4+cos2(x))(1+cos2(x))dx

Now use the substitution u=cos(x) so that dx=−1sin(x)du and we have

I=−∫1(4+u2)(1+u2)du

Split into partial fractions:

I=13∫1u2+4−1u2+1du

Now use standard results

I=13(12arctan(u2)−arctan(u))+C

Finally, undo the substitution to get

I=16arctan(12cos(x))−13arctan(cos(x))+C