# Integrating $\int\frac{5}{\ 16 + 9\cos^2(x)}\,dx$

I am trying to integrate the following:

$$\int\frac{5}{\ 16 + 9\cos^2(x)}\,dx$$

I have applied the following substitution:

$$x = \tan^{-1}u$$

I have simplified the denominator through using the following trig identity:

$$\cos^{2}x = 1/(1 + \tan^{2}x)$$

$$16 + 9(1/(1 + \tan^{2}x))$$

$$\frac{16(1 + tan^{2}x) + 9}{\ 1 + \tan^{2}x}\$$

$$= \frac{25 + 16\tan^{2}x}{\ 1 + \tan^{2}x}\$$

Substituting the above into the denominator I get:

$$5\int\frac{1 + \tan^{2}x}{\ 25 + 16\tan^2(x)}\,dx$$

However, I know that the result of the above substitution should be:

$$5\int\frac{1}{\ 16u^{2} + 25}\,du$$

I am very close to this result except for the fact that the numerator in my integral is $$1 + \tan^{2}x$$ instead of 1.

I am not sure how I can get rid of the $$\tan^{2}x$$ in my numerator. Any insights are appreciated.

Notice, you should use $$1+\tan^2x=\sec^2x$$.

You can get to that integral as follows

$$=5\int\frac{1+\tan^2x}{\ 16\tan^2 x + 25}\,dx$$ $$=5\int\frac{\sec^2x}{\ 16\tan^2 x + 25}\,dx$$ Let $$\tan x=u\implies \sec^2x\ dx=du$$ $$=5\int\frac{du}{16u^2 + 25}$$

Note that $$\cos(x)=\frac{1}{\sec(x)}$$ and $$\sec^2(x)=\tan^2(x)+1$$ so the integral becomes:

$$\int\sec^2(x)\times\frac{1}{16\tan^2(x)+25}$$

Substitute $$u=\tan(x)$$ so that $$dx=\frac{1}{sec^2(x)}du$$ and the integral becomes:

$$\int\frac{1}{16u^2+25}$$

Now substitute $$v=\frac{4u}{5}$$ so $$du=\frac{5}{4}dv$$ and the integral becomes:

$$\frac{1}{20}\int\frac{1}{v^2+1}$$ $$=\frac{\arctan(v)}{20}$$

Substitute $$v$$ back to get:

$$=\frac{\arctan(\frac{4u}{5})}{20}$$

Then substitute u back to get:

$$=\frac{\arctan(\frac{4\tan(x)}{5})}{20}$$

Note that when you got to

$$5\int{1+\tan^2x\over25+16\tan^2x}dx$$

you haven't yet used the substitution $$x=\arctan u$$. If you notice $$x=\arctan u$$ implies $$dx={du\over1+u^2}$$, you see immediately that

$${1+\tan^2x\over25+16\tan^2x}dx={1+u^2\over25+16u^2}\cdot{du\over1+u^2}={1\over25+16u^2}du$$

so the result of the subsitution is indeed

$$5\int{1\over25+16u^2}du$$