Homework help: Evaluate the following integrals!

[Robd0ggie]

1) integrate 1/Sqrt(25-9x^2) dx


2) integrate cos x/(1+sin^2 x) dx from 0 to pi/3



I attempted using U-substitution, but it doesn't seem to make sense. After reviewing my professor's notes and my Calculus, i cannot find a method to solve these last two questions of my homework. I don't want to give up, I'm sitting in class with 89% and i want to pass with an A.

Please show your steps, so that i may also learn.

Any help is appreciated, Thank you!

 

[Deleted member 4993]

1) integrate 1/Sqrt(25-9x^2) dx


2) integrate cos x/(1+sin^2 x) dx from 0 to pi/3



I attempted using U-substitution, but it doesn't seem to make sense. After reviewing my professor's notes and my Calculus, i cannot find a method to solve these last two questions of my homework. I don't want to give up, I'm sitting in class with 89% and i want to pass with an A.

Please show your steps, so that i may also learn.

Any help is appreciated, Thank you!

Substitution should work!!

For problem (1) substitute

$\displaystyle u \ = \ 3x$

For problem (2) substitute

$\displaystyle u \ = \ sin(x)$

Please share your work with us indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[galactus]

1) integrate 1/Sqrt(25-9x^2) dx

Let \(\displaystyle x=\frac{5}{3}sin\theta, \;\ dx=\frac{5}{3}cos\thetad\theta\)

Upon making the subs, it whittles down to practically nothing.

2) $\displaystyle \int_{0}^{\frac{\pi}{3}} \frac{cos(x)}{1+sin^{2}(x)}dx$

You can use a u sub here. Let $\displaystyle u=sin(x), \;\ du=cos(x)dx$

After making the subs, you should recognize something related to an inverse trig function.

 

[soroban]

Hello, Robd0ggie!

$\displaystyle \text{1) Integrate: }\;\int\frac{dx}{\sqrt{25-9x^2}}$

This has the form of the arcsine integeral.

. . $\displaystyle \int\frac{du}{\sqrt{a^2-u^2}} \;=\;\sin^{-1}\!\left(\frac{u}{a}\right) + C$


$\displaystyle \text{We have: }\:a = 5,\;u = 3x \quad\Rightarrow\quad du = 3\,dx \quad\Rightarrow\quad dx = \tfrac{1}{3}du$

$\displaystyle \text{Our integral becomes: }\:\int\frac{\frac{1}{3}du}{\sqrt{a^2 - u^2}} \;=\;\tfrac{1}{3}\int\frac{du}{\sqrt{a^2-u^2}} \;=\;\tfrac{1}{3}\sin^{-1}\left(\tfrac{u}{a}\right) + C$

$\displaystyle \text{Back-substitute: }\:\tfrac{1}{3}\sin^{-1}\left(\tfrac{3x}{5}\right) + C$

$\displaystyle \text{2) Evaluate: }\:\int^{\frac{\pi}{3}}_0 \frac{\cos x\,dx}{1+\sin^2 x}$

This has the form of the arctangent integral.

. . $\displaystyle \int \frac{du}{a^2 + u^2} \;=\;\frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right) + C$


$\displaystyle \text{We have: }\:a = 1,\;u = \sin x,\;du = \cos x\,dx$

$\displaystyle \text{The integral becomes: }\:\int \frac{du}{a^2+u^2} \:=\:\frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right) + C$

$\displaystyle \text{Back-substitute: }\:\frac{1}{1}\tan^{-1}\left(\frac{\sin x}{1}\right) + C \;=\;\tan^{-1}(\sin x) + C$

$\displaystyle \text{Evaluate: }\:\tan^{-1}(\sin x)\,\bigg]^{\frac{\pi}{3}}_0 \;=\; \tan^{-1}(\sin\tfrac{\pi}{3}) - \tan^{-1}(\sin 0)$

. . $\displaystyle = \;\tan^{-1}\left(\tfrac{\sqrt{3}}{2}\right) - \tan^{-1}(0) \;=\;\tan^{-1}\left(\tfrac{\sqrt{3}}{2}\right) \;\approx\;0.713724379$