Indefinite integral of $\frac{\sec^2x}{(\sec x+\tan x)^\frac{9}{2}}$

Question

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

My approach:

Since it is easy to evaluate $\int{\sec^2x}$ , integration by parts seems like a viable option.

Let $$I_n=\int{\frac{\sec^2x}{(\sec x+\tan x)^\frac{9}{2}}}$$

$$I_n=\frac{\tan x}{(\sec x+\tan x)^\frac{9}{2}} + \frac{9}{2}\int{\frac{\sec x \tan x}{(\sec x+\tan x)^\frac{9}{2}}dx}$$

Evaluating the new integral again using by parts yields

$$\frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}}+\frac{9}{2}\int{\frac{\sec^2x}{(\sec x+\tan x)^\frac{9}{2}}\,dx}$$

$$=\frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}} + \frac{9}{2} I_n$$

Plugging it back, we obtain $$I_n=\frac{\tan x}{(\sec x+\tan x)^\frac{9}{2}} + \frac{9}{2}\frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}} + \frac{81}{4}I_n $$

$$\frac{-77}{4}I_n=\frac{\tan x}{(\sec x+\tan x)^\frac{9}{2}} + \frac{9}{2}\frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}}$$

This obviously doesn't match with bprp's answer. Help!

Edit: How do I convert my answer to the answer obtained by him, if mine is correct

https://math.stackexchange.com/questions/1054719/evaluating-int-frac-sec2-x-sec-x-tan-x-9-2-mathrm-dx — lab bhattacharjee, Aug 05 '20 at 10:11
Does this answer your question? Evaluating $\int \frac{\sec^2 x}{(\sec x + \tan x )^{{9}/{2}}},\mathrm dx$ — Alessio K, Aug 05 '20 at 10:12
I have edited the question asking for help to transform my result — DatBoi, Aug 05 '20 at 10:21

Math Lover · Accepted Answer · 2020-08-05T11:21:11.423

$-77I_n=4 \times \frac{\tan x}{(\sec x+\tan x)^\frac{9}{2}} + 18 \times \frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}}$

$-77I_n= 11 \times \frac{\tan x}{(\sec x+\tan x)^\frac{9}{2}} - 7 \times \frac{\tan x}{(\sec x+\tan x)^\frac{9}{2}} + 11 \times \frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}} + 7 \times \frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}}$

$-77I_n= 11 \times \frac{\sec x+\tan x}{(\sec x+\tan x)^\frac{9}{2}} + 7 \times \frac{\sec x - \tan x}{(\sec x+\tan x)^\frac{9}{2}}$

First part is straightforward. The second part can be simplified multiplying both numerator and denominator by $(\sec x+\tan x)$. We know, $(\sec x+\tan x) (\sec x-\tan x) = 1$.

You can take it from here.

score 0 · Answer 2 · answered Oct 05 '20 at 17:40

You missed a simplification after second step:

$$I=\frac{\tan x}{(\sec x+\tan x)}\frac{9}{2}+\frac{9}{2}\int \frac{\sec x \tan x}{(\sec x+\tan x)^{\frac92}}dx$$

Now, take the original expression for $I$

$$ I = \int \frac{\sec^2 x}{(\sec x + \tan x)^{\frac92}}$$

Add this $ \frac{9}{2} $ times this to previous expression

$$ \frac{11}{2} I = \frac{\tan x}{(\sec x+\tan x)}\frac{9}{2}+\frac{9}{2}\int \frac{\sec x \tan x + \sec^2 x}{(\sec x+\tan x)^{\frac92}}dx$$

Consider the second integral:

$$ J = \int \frac{\sec x \tan x + \sec^2 x}{(\sec x+\tan x)^{\frac92}}dx$$ Substitute $ \sec x + \tan x = t$

$$ J = \int t^{-\frac92} dt$$

Done!

Indefinite integral of $\frac{\sec^2x}{(\sec x+\tan x)^\frac{9}{2}}$

2 Answers2