What is the integral of cos(x)/(5+sin^2(x))dx?

2 Answers
May 6, 2018

int \frac{cos x}{5 + sin^2 x}\ d x = \frac{1}{\sqrt{5}} tan^{-1} \frac{sin x}{\sqrt{5}} + C

Explanation:

The cosine in the enumerator almost urges us to view sin x as an inner function (with derivative cos x). Therefore we can write the integral as

[ int \frac{1}{5 + t^2}\ d t ]_{t = sin x} =

= [ \frac{1}{5} int \frac{1}{1 + \frac{t^2}{5}}\ d t ]_{t = sin x} =

= [ \frac{1}{5} int \frac{1}{1 + ( \frac{t}{\sqrt{5}} )^2}\ d t ]_{t = sin x}.

By viewing t / \sqrt{5} as yet another inner function, with derivative 1 / \sqrt{5}, this is the same as

= [ \frac{1}{\sqrt{5}} int \frac{1}{1 + u ^2}\ d u ]_{u = \frac{sin x}{\sqrt{5}}}.

Now we are pretty much done, however, since the remaining integrand is the derivative of tan^{-1}. Therefore, we have

[ \frac{1}{\sqrt{5}} \tan^{-1} u + C ]_{u = \frac{sin x}{\sqrt{5}}} =

= \frac{1}{\sqrt{5}} tan^{-1} \frac{sin x}{\sqrt{5}} + C.

May 6, 2018

The answer is =1/sqrt5arctan(sinx/sqrt5)+C

Explanation:

We need

int(dx)/(1+x^2)=arctanx+C

Perform the substitution

u=sinx/sqrt5, =>, du=(cosxdx)/sqrt5

The integral is

int(cosxdx)/(5+sin^2x)=1/5int(cosxdx)/(1+(sinx/sqrt5)^2)

=sqrt5/5int(du)/(1+u^2)

=1/sqrt5*arctanu

=1/sqrt5arctan(sinx/sqrt5)+C