1. ## HARD Integration

hey, last week we had a homework with f**** 26 Integration problems, i killed my self and couldnt solve this 9, can anyone help with any ?sort of teach how to do

1. tan ^ 3 x sec x dx
2. sec ^ 4 x dx
3. [ dx / 2 - sin ^ 2 x ] dx
4. [ sin ^ 3 x / root over cos x ] dx
5. dx / 3 + 5 cos x
6. [ sin x / 1 + sin x ] dx
7. [ root over x / 1 + 4 root over x ^ 3 ] dx in here 4 is "on" the root over sign, i dont know how to say that in english, ****
8. [ root over ( 3 + 2x ) / x ] dx
9. dx / root over ( x + 1 ) + root over x

thankssssssssss

2. Please watch the language. This is a square house.

Anyway, the first one is not that bad.

$\int tan^{3}(x)sec(x)dx$

$\int tan(x)sec(x)(sec^{2}(x)-1)dx$

Let $u=sec(x), \;\ du=sec(x)tan(x)dx$

$\int (u^{2}-1)du$

Go for it.

3. Hi,
Originally Posted by masterjoint
9. dx / root over ( x + 1 ) + root over x
$
\frac{1}{\sqrt{x+1}+\sqrt{x}}=\frac{1}{\sqrt{x+1}+ \sqrt{x}}\cdot\frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1}-\sqrt{x}}=\frac{\sqrt{x+1}-\sqrt{x}}{x+1-x}=\sqrt{x+1}-\sqrt{x}$

...

4. 4. $\int\frac{sin ^ {3} (x)}{\sqrt{ cos (x)}} dx$
Let $u=cos(x), \;\ -du=sin(x)dx$

Make the subs and get:

$\int\left[u^{\frac{-1}{2}}-u^{\frac{3}{2}}\right]du$

5. Hello,

8. $\int \frac{\sqrt{3+2x}}{x} ~dx$
Substitute $t=\sqrt{3+2x}$

hence $t^2=3+2x \implies x=\frac{t^2-3}{2}$

So $dx=t dt$

The integral is now :

$\int \frac{t}{x} ~ t ~ dt=\int \frac{t^2}{\frac{t^2-3}{2}} ~ dt=2 \int \frac{t^2}{t^2-3} ~ dt=2 \int 1+\frac{3}{t^2-3} ~ dt=2t+6 \int \frac{1}{t^2-3} ~ dt$

now you can use partial fraction decomposition : $t^2-3=(t-\sqrt{3})(t+\sqrt{3})$

6. Originally Posted by masterjoint
5. dx / 3 + 5 cos x
Substitute $t=\tan(x/2)$ :

$\int\frac{1}{3+5\cos x }\,\mathrm{d}x=\int \frac{1}{3+5\cdot\frac{1-t^2}{1+t^2}}\cdot \frac{2\mathrm{d}t}{1+t^2}=\int\frac{\mathrm{d}t}{ 4-t^2}$

Can you take it from here ?

7. 6. $\int \frac{\sin(x)}{1+\sin(x)} ~dx$
$=\int 1-\frac{1}{1+\sin(x)} ~ dx$

$=\int ~ dx -\int \frac{1}{1+\sin(x)} ~dx$

use Weierstrass substitution $t=\tan \frac x2$ :

$=x-\int \frac{1}{1+\frac{2t}{1+t^2}} \cdot \frac{2}{1+t^2} ~ dt$

$=x-2 \int \frac{1}{1+2t+t^2} ~ dt$

noting that $1+2t+t^2=(t+1)^2$, it is easy from now

8. Originally Posted by masterjoint

6. [ sin x / 1 + sin x ] dx
$\int{\frac{\sin x}{1+\sin x}\,dx}=\int{\frac{\sin x(1-\sin x)}{\cos ^{2}x}\,dx}=\int{\tan x\sec x\,dx}-\int{\tan ^{2}x\,dx},$ and the last integral is easy to solve since $1+\tan^2x=\sec^2x.$

9. ## thank you !

thanks alot!!!!!,. really ~!