# Thread: integrate this function :S

1. ## integrate this function :S

Dont know where to start..

Integrate

< avoid this question >(a^px)*(b^px)

also integrate this function

(sinx + cosx)/(sin^4x + cos^2x)

please dont send me a wolfram alpha link.... -_-
can we do this question witout using the universal substitution tanx/2 = t ?

2. Originally Posted by ice_syncer
Dont know where to start..

Integrate

(a^px)*(b^px)
$a^{px} \cdot b^{px} = (ab)^{px}$

let $c = ab$ ...

$\displaystyle \int c^{px} \, dx = \frac{c^{px}}{p \ln{c}} + C = \frac{(ab)^{px}}{p \ln(ab)}+C$

3. sorry thats the wrong question

actually it is integrate (a^px)*(b^qx)
sorryy

4. Originally Posted by ice_syncer
sorry thats the wrong question

actually it is integrate (a^px)*(b^qx)
sorryy
$a^{px} \cdot b^{qx} = (a^p b^q)^x$

since a, b, p, and q are all constants ...

let $c = a^p b^q$

$\displaystyle \int c^x \, dx = \frac{c^x}{\ln{c}} + C
$

back substitute ...

$\displaystyle \frac{(a^p b^q)^x}{\ln(a^p b^q)} + C$

5. oh yeah, ok. thanks : )

6. Originally Posted by ice_syncer
Dont know where to start..
Integrate (a^px)*(b^px)

(i) If you mean $\int (a^px)(b^px)dx$, then:

$\displaystyle\int (a^px)(b^px)dx=(ab)^{p}\int x^2dx=\dfrac{(ab)^px^3}{3}+C$

(ii) If you mean $\int (a^{px})(b^{px})dx$, then:

$\displaystyle\int (a^{px})(b^{px})dx=\displaystyle\int (ab)^{px}dx=\dfrac{1}{p\log (ab)}(ab)^{px}+C$

Regards.

Edited: Sorry, when I submitted my answer, I didn't see the previous posts. Besides, I read $b^{px}$ instead of $b^{qx}$ .

7. thanks

8. the next question is hard for me becauseof complicated partial fractions after substituting, actually you get surd like stuff

9. Originally Posted by ice_syncer
the next question is hard for me becauseof complicated partial fractions after substituting, actually you get surd like stuff
I would guess so ... Wolfram gives a really ugly antiderivative in terms of an inverse hyperbolic tangent and inverse tangent.

10. someone help me with this