# Thread: [SOLVED] definite intergral problem

1. ## [SOLVED] definite intergral problem

Hi - I'd appreciate it if someone could point out where i'm going wrong on this one:

Intergrate within the limits of pi/2 and 0

(sinx)^3 * (cosx)^2 dx

Ok, use the identity (sinx)^2 = 1 - (cosx)^2

and I get sinx(cosx)^2 - sinx(cosx)^4

Intergrate and i get:

-1/3(cosx)^3 + 1/5(cosx)^5

Evaluate within the limits and I get zero as cos pi/2 equals zero.

This isnt the answer i get on an itegration calculator???

thanks D

2. Originally Posted by dojo
Hi - I'd appreciate it if someone could point out where i'm going wrong on this one:

Intergrate within the limits of pi/2 and 0

(sinx)^3 * (cosx)^2 dx

Ok, use the identity (sinx)^2 = 1 - (cosx)^2

and I get sinx(cosx)^2 - sinx(cosx)^4

Intergrate and i get:

-1/3(cosx)^3 + 1/5(cosx)^5

Evaluate within the limits and I get zero as cos pi/2 equals zero.

This isnt the answer i get on an itegration calculator???

thanks D

But $\displaystyle \cos 0=1$ !! So the answer must be:

$\displaystyle \frac{1}{3}[\cos^3(\pi\slash 2)-\cos^3(0)]+\frac{1}{5}[\cos^5(\pi\slash 2)-\cos^5(0)]=-\frac{1}{3}(-1)+\frac{1}{5}(-1)=\frac{1}{3}-\frac{1}{5}=\frac{2}{15}$

Tonio

3. Thanks - Schoolboy error . I've had my nose in my books too long today!

4. Originally Posted by dojo
Hi - I'd appreciate it if someone could point out where i'm going wrong on this one:

Intergrate within the limits of pi/2 and 0

(sinx)^3 * (cosx)^2 dx

Ok, use the identity (sinx)^2 = 1 - (cosx)^2

and I get sinx(cosx)^2 - sinx(cosx)^4

Intergrate and i get:

-1/3(cosx)^3 + 1/5(cosx)^5

Evaluate within the limits and I get zero as cos pi/2 equals zero.

This isnt the answer i get on an itegration calculator???

thanks D
I will show you the steps involved. I hope you can plug in the limits and calculate:

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

= $\displaystyle \int sinx \times cos^{2}x \times sin^{2}x dx$

= $\displaystyle \int sinx \times cos^{2}x (1-cos^{2}x) dx$

Now use substitution rule:

let $\displaystyle u = cosx ,$ therefore $\displaystyle du = -sinx dx$ that gives $\displaystyle dx = \frac {-du}{sinx}$

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

= $\displaystyle - \int ({u^2}-{u^4}) du$

= $\displaystyle \int {u^4}du - \int {u^2}du$

= $\displaystyle \frac{u^5}{5} - \frac{u^3}{3}$

substitute $\displaystyle u = cosx$ to get

= $\displaystyle [\frac {cos^{5}x}{5}] - [\frac{cos^{3}x}{3}]$

then evaluate these fractions with the limits you have...