# Math Help - Please walk me through this integral

1. ## Please walk me through this integral

(sin (x))(tan^2 (x))

Kim

2. Originally Posted by Kim Nu
(sin (x))(tan^2 (x))

Kim

There may be a faster way but this is how I would do it.
rewirte the integrand as

$\sin(x)\cdot \frac{\sin^{2}(x)}{\cos^2(x)}=\sin^{3}(x)\sec^{2}( x)$

Now we will integrate by parts with

$u=\sin^{3}(x) \to du = 3\sin^{2}(x)\cos(x)$

$dv=\sec^{2}(x) \to v=\tan(x)$

$
\sin^{3}(x)\tan(x)-\int 3\sin^{2}(x)\cos(x)\tan(x)dx
$

rewrite the integrand as

$
\sin^{3}(x)\tan(x)-\int 3\sin^{3}(x)dx = \sin^{3}(x)\tan(x)-\int 3\sin(x)(1-\cos^2(x))dx=$

$\sin^{3}(x)\tan(x)-3\int \sin(x)dx+3\int \sin(x)\cos^2(x))dx
$

The last integral can be done with a u sub u=cos(x)

and we finally get

$\sin^{3}(x)\tan(x)+3\cos(x)-\cos^3(x)
$

Maybe there is a faster way....

Anyone...

I hope this helps

3. Originally Posted by Kim Nu
(sin (x))(tan^2 (x))

Kim

Okay this works even faster

$\int \sin(x)\tan^{2}(x)$

by parts again

$u=\tan^{2}(x) \to du=2\tan(x)\sec^{2}(x)dx$

$dv=\sin(x) \to v= -\cos(x)$

Now we get

$-\cos(x)\tan^{2}(x)-\int(-\cos(x))(2\tan(x)\sec^{2}(x))dx$

Cleaning up a bit gives

$-\cos(x)\tan^{2}(x)+2\int\frac{\sin(x)}{\cos^{2}(x) }dx$

Now just let u=cos(x) to finish

So we end up with

$-\cos(x)\tan^{2}(x)+\frac{2}{\cos(x)}=$

$-\cos(x)\tan^{2}(x)+2\sec(x)+C$

4. This approach uses only one substitution:

$\int\sin x \tan^2 x~dx$

$\int \frac{\sin^3 x}{\cos^2 x}~dx$

$\int \frac{\sin x (1-\cos^2 x)}{\cos^2 x}~dx$

$\int \frac{\sin x}{\cos^2 x} - \sin x~dx$

$\int \frac{\sin x}{\cos^2 x}~dx + \cos x + C$

Let $u = \cos x$, then $du = -\sin x~dx$

$-\int \frac{du}{u^2} + \cos x + C$

$\frac{1}{u} + \cos x + C$

$\frac{1}{\cos x} +\cos x + C$

5. Hey thanks everyone.

Thanks, Kim

6. Originally Posted by Kim Nu
Hey thanks everyone.

Thanks, Kim
This integral doesn't have an easy solution, it involves the use of hyperbolic sine or cosine functions. Do you know them?

7. Originally Posted by Kim Nu
Hey thanks everyone.

Thanks, Kim
Let $u=\sin(x)$ then: $
\int {\frac{{\sin ^2 \left( x \right)}}
{{\cos ^3 \left( x \right)}}dx} = \int {\frac{{u^2 }}
{{\left( {1 - u^2 } \right)^2 }}du}
$

note that: $
\int {\frac{{u^2 }}
{{\left( {1 - u^2 } \right)^2 }}du} = \tfrac{1}
{2}\int {u \cdot \left( {\frac{1}
{{1 - u^2 }}} \right)^\prime du}
$
(1)

Apply parts in (1)

And remember that: $\frac{1}{(1-u^2)}=\frac{1}{2}\cdot{\left(\frac{1}{1-u}+\frac{1}{1+u}\right)}$

8. A slight improvement on wingless's solution . . .

This approach uses only one substitution:

$\int\sin x \tan^2\!x\,dx \;=\;\int \frac{\sin^3\!x}{\cos^2\!x}\,dx \;=\;\int \frac{\sin x (1-\cos^2\!x)}{\cos^2\!x}\,dx \;=$ . $\int\left(\frac{\sin x}{\cos^2\!x} - \sin x\right)\,dx$

We have: . $\int\left(\frac{1}{\cos x}\!\cdot\!\frac{\sin x}{\cos x} - \sin x\right)\,dx$

. . . . . . $\;\;=\;\;\int(\sec x\tan x - \sin x)\,dx \;=\;\boxed{\sec x + \cos x + C}$

9. Hello, Kim!

Are you familiar with this formula?

. . $\int\sec^3\!x\,dx \;=\;\frac{1}{2}\bigg[\sec x\tan x + \ln|\sec x + \tan x|\bigg] + C$

How about this one: . $\int\frac{\tan^2\!x}{\cos x}\,dx$

We have: . $\int\sec x\tan^2\!x\,dx \;=\;\int\sec x(\sec^2\!x - 1)\,dx \;=\;\int\left(\sec^3\!x - \sec x\right)dx$

Then: . $\frac{1}{2}\bigg[\sec x\tan x + \ln|\sec x + \tan x|\bigg] - \ln|\sec x + \tan x| + C$

. . . $= \;\frac{1}{2}\sec x\tan x - \frac{1}{2}\ln|\sec x + \tan x| + C$