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I am unable to solve the following integral: $$\int \frac{\sec^{2}{x}}{(\sec{x} + \tan{x})^{5/2}} \ \mathrm dx.$$

Tried putting $t=\tan{x}$ so that numerator has $\sec^{2}{x}$ but that doesn't help.

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First let $y=\tan{x}$ so the integral becomes

$$\int \frac{dy}{\left (y+\sqrt{1+y^2}\right)^{5/2}}$$

Now let $u = y+\sqrt{1+y^2}$; then I find that

$$y=\frac12 \left (u-\frac{1}{u}\right ) \implies dy = \frac12 \left (1+\frac{1}{u^2} \right )$$

Then the integral is

$$\frac12 \int du \left (u^{-5/2}+u^{-9/2} \right ) = -\frac13 u^{-3/2}-\frac17 u^{-7/2}+C $$


$$\int dx \frac{\sec^2{x}}{(\sec{x}+\tan{x})^{5/2}} = -\frac13 (\sec{x}+\tan{x})^{-3/2} - \frac17 (\sec{x}+\tan{x})^{-7/2} + C $$

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Here are some hints:

  • Put $t = \sec{x} + \tan{x}$.

  • Use the identity $\sec^{2}{x} - \tan^{2}{x} = 1$, to get the value of $\sec{x}$ in terms of $t$.

Complete Solution:

$t = \sec{x} + \tan{x} \Rightarrow dt = \sec{x} \cdot (\sec{x}+\tan{x}) dx \Rightarrow dt = \Bigl(t+\frac{1}{t}\Bigr)\cdot t \ dx$. By following the above you get $t+\frac{1}{t} = \sec{x}$. So your integral becomes \begin{align*} \int \frac{\sec^{2}{x}}{(\sec{x}+\tan{x})^{5/2}}\ dx &= \int \frac{1}{t^{5/2}} \times \biggl(t+\frac{1}{t}\biggr)^{2} \times \frac{dt}{(t+\frac{1}{t}) \cdot t}\end{align*}

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This appears to be different from the previous substitutions, and perhaps a little more intuitive and showing more intermediate steps (e.g. we could have skipped straight to $(3)$ using $w=(\sec(x)-\tan(x))^2$). $$ \begin{align} &\int\frac{\sec^2(x)\,\mathrm{d}x}{(\sec(x)+\tan(x))^{5/2}}\\ &=\int\frac{\cos(x)^{-1/2}\,\mathrm{d}\sin(x)}{(1+\sin(x))^{5/2}}\\ &=\int\frac{(1-u^2)^{-1/4}\,\mathrm{d}u}{(1+u)^{5/2}}\tag{1}\\ &=\int\frac{(1-u)^{-1/4}\,\mathrm{d}u}{(1+u)^{11/4}}\\ &=\int(1-2v)^{-1/4}(1-v)\,\mathrm{d}v\tag{2}\\ &=-\frac14\int w^{-1/4}(1+w)\mathrm{d}w\tag{3}\\ &=-\frac13w^{3/4}-\frac17w^{7/4}+C\\ &=-\frac13\left(\sec(x)-\tan(x)\right)^{3/2} -\frac17\left(\sec(x)-\tan(x)\right)^{7/2}+C\tag{4} \end{align} $$ $(1)$: $u=\sin(x)\vphantom{\left(\frac12\right)}$
$(2)$: $u=\frac{v}{1-v}\quad \mathrm{d}u=\frac{\mathrm{d}v}{(1-v)^2}\quad 1+u=\frac1{1-v}\quad 1-u=\frac{1-2v}{1-v}$
$(3)$: $1-2v=w\quad 1-v=\frac{1+w}{2}\quad \mathrm{d}v=-\frac12\mathrm{d}w$
$(4)$: $w=\frac{1-\sin(x)}{1+\sin(x)}=\left(\frac{1-\sin(x)}{\cos(x)}\right)^2=(\sec(x)-\tan(x))^2$

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A Generalization for

Setting $\displaystyle x=\frac\pi2-2y$

$\displaystyle\implies(i) dx=-2dy$

and $\displaystyle(ii)\sec x=\sec\left(\frac\pi2-2y\right)=\csc2y$ and similarly, $\displaystyle\tan\left(\frac\pi2-2y\right)=\cot2y$

$$I=\int \frac{\sec^mx}{(\sec x + \tan x)^n}dx=-\int\frac{\csc^m2y}{(\csc2y+\cot2y)^n}2dy$$


Using the Weierstrass substitution, $$I=-\int\frac{(1+\tan^2y)^{m-1}\sec^2ydy}{2^{m-1}(\tan y)^{m-n}}$$

Setting $\tan y=u,$ $$I=-\int\frac{(1+u^2)^{m-1}du}{2^{m-1}u^{m-n}}$$

This can be solved easily using Binomial Expansion

if $(i)$ integer $m\ge1$

and if $(ii)n$ is not an integer(so will be $m-n$ as $m$ is an integer)

we need to add one assumption $\displaystyle \sin y,\cos y>0$ i..e, $y$ lie sin the first quadrant, else we need to handle the signs of $(\sin2y)^{m-n}, (1+\cos2y)^n$ carefully.

Now $\displaystyle u=\tan y=\tan\left(\frac\pi4-\frac x2\right)=\frac{1-\tan\frac x2}{1+\tan\frac x2}=\frac{\cos\frac x2-\sin\frac x2}{\cos\frac x2+\sin\frac x2}$

$\displaystyle=\frac{\left(\cos\frac x2-\sin\frac x2\right)\left(\cos\frac x2+\sin\frac x2\right)}{\left(\cos\frac x2+\sin\frac x2\right)\left(\cos\frac x2+\sin\frac x2\right)}=\frac{\cos^2\frac x2-\sin^2\frac x2}{1+2\sin\frac x2\cos\frac x2}$ $\displaystyle=\frac{\cos x}{1+\sin x}=\frac1{\sec x+\tan x}=\sec x-\tan x$

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Take $t=\sec(x)-\tan(x)$; so; $$1/t=\sec(x)+\tan(x)$$ $$(1/t-t)/2=\tan(x)$$ $$(-1/t^2-1)/2 \,dt=\sec^2(x)dx\,$$ Now just substitute & answer will come

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Welcome to Math.SE. I edited your post. Please see math notation guide for formatting tips. – user147263 Sep 3 '14 at 15:27

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