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Thread: Basic Integration

  1. #1
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    Basic Integration

    I've been studying for a integration and volumes test. I can do most of them but these few are giving me headaches. You don't have to solve them for me, just tell me what I need to do. Any help is appreciated.

    1)
    $\displaystyle \int \sqrt{16+x^2} \,dx$

    I know I need to substitute $\displaystyle x = 4\,tan\,\theta \Rightarrow dx = 4\,sec^2\theta \,d\theta$
    $\displaystyle \int 4sec\theta \bullet 4sec^2\theta \,d\theta$
    =$\displaystyle \int 16 sec^3\theta\,d\theta$

    I don't know how to go on from here.
    The answer is $\displaystyle \frac{1}{2}x\sqrt{16+x^2} + 8 ln (x+\sqrt{16+x^2})$

    2)
    $\displaystyle \int \frac{1}{x^2\sqrt{x-1}}dx$

    I've tried letting $\displaystyle u= \sqrt{x-1}$, but got nowhere from that.

    3)
    $\displaystyle \int_0^{\frac{\pi}{2}} \sqrt{1+sin\,2x}\, dx$

    no clue
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  2. #2
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    Quote Originally Posted by Gusbob View Post
    2)$\displaystyle \int \frac{1}{x^2\sqrt{x-1}}dx$

    I've tried letting $\displaystyle u= \sqrt{x-1}$, but got nowhere from that.
    Try $\displaystyle x = sec^2 \theta$

    Quote Originally Posted by Gusbob View Post
    3)$\displaystyle \int_0^{\frac{\pi}{2}} \sqrt{1+sin\,2x}\, dx$

    no clue
    Did you observe $\displaystyle \sqrt{1+sin\,2x} = |\cos x + \sin x|$ ?
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  3. #3
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    1.
    Let$\displaystyle u=\sin(x)$

    $\displaystyle \int \frac{dx}{\cos^3(x)}=\int \frac{\cos(x)}{\cos^4(x)}dx=\int \frac{du}{(1-u^2)^2}$

    $\displaystyle \frac{1}{(1-u^2)}=\frac{1}{2}\cdot{\left(\frac{1}{1-u}+\frac{1}{1+u}\right)}$ (1)

    square (1) and apply (1) again with what you get

    2.
    After that sub you get $\displaystyle 2\int\frac{du}{(u^2+1)^2}$

    Let: $\displaystyle z=\arctan(u)$ and remember that $\displaystyle \frac{1}{\tan^2(z)+1}=\cos^2(z)$
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    Did you observe $\displaystyle \sqrt{1+sin\,2x} = |\cos x + \sin x|$ ?
    How does that work?
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    Hello,

    Quote Originally Posted by Gusbob View Post
    How does that work?
    $\displaystyle 1=\cos^2x+\sin^2x$

    $\displaystyle \sin 2x=2 \cos x \sin x$

    $\displaystyle \implies 1+\sin 2x=\cos^2x+\sin^2x+2 \cos x \sin x=(\cos x+\sin x)^2$
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  6. #6
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    Quote Originally Posted by Gusbob View Post
    I've been studying for a integration and volumes test. I can do most of them but these few are giving me headaches. You don't have to solve them for me, just tell me what I need to do. Any help is appreciated.

    1)
    $\displaystyle \int \sqrt{16+x^2} \,dx$

    I know I need to substitute $\displaystyle x = 4\,tan\,\theta \Rightarrow dx = 4\,sec^2\theta \,d\theta$
    $\displaystyle \int 4sec\theta \bullet 4sec^2\theta \,d\theta$
    =$\displaystyle \int 16 sec^3\theta\,d\theta$

    I don't know how to go on from here.
    The answer is $\displaystyle \frac{1}{2}x\sqrt{16+x^2} + 8 ln (x+\sqrt{16+x^2})$

    2)
    $\displaystyle \int \frac{1}{x^2\sqrt{x-1}}dx$

    I've tried letting $\displaystyle u= \sqrt{x-1}$, but got nowhere from that.

    3)
    $\displaystyle \int_0^{\frac{\pi}{2}} \sqrt{1+sin\,2x}\, dx$

    no clue
    For the first one by trig sub we get

    Let $\displaystyle x=4\tan(\theta)\Rightarrow{\theta=arctan\bigg(\fra c{x}{4}\bigg)}$

    This means that $\displaystyle dx=16\sec^2(\theta)$

    Imputting this we get

    $\displaystyle \int\sqrt{16+16\tan^2(\theta)}\sec^2(\theta)d\thet a=4\int\sec^3(\theta)d\theta$

    Now to integrate that thing

    Ok so we do this

    remember that $\displaystyle \sec^3(x)=(1+\tan^2(x))\sec(x)=\sec(x)+\sec(x)\tan ^2(x)$

    So subbing this in remember waht this is actually equal to we get

    $\displaystyle \int\sec(\theta)+\sec(\theta)\tan^2(\theta)d\theta =\ln|\sec(\theta)+\tan(\theta)|+\int\sec(\theta)\t an^2(\theta)d\theta$

    Now evaluating the integral within their we let

    $\displaystyle dv=\sec(\theta)\tan(\theta)d\theta$

    and let $\displaystyle u=\tan(\theta)$

    So $\displaystyle v=sec(\theta)$

    and $\displaystyle du=\sec^2(\theta)d\theta$

    Applying parts we get

    $\displaystyle \int\sec(\theta)\tan^2(\theta)d\theta=\sec(\theta) \tan(\theta)-\int\sec^3(\theta)d\theta$

    Putting this back into the integral we get

    $\displaystyle \int\sec(\theta)+\sec(\theta)\tan^2(\theta)d\theta =\int\sec^3(\theta)d\theta=$$\displaystyle \ln|\sec(\theta)+\tan(\theta)|+\sec(\theta)\tan(\t heta)-\int\sec^3(\theta)d\theta$$\displaystyle \Rightarrow{2\int\sec^3(\theta)d\theta=\ln|\sec(\t heta)+\tan(\theta)|+\sec(\theta)\tan(\theta)}$

    Dividing both sides by two we get

    $\displaystyle 16\int\sec^3(\theta)d\theta=2\bigg[\ln|\sec(\theta)+\tan(\theta)|+\sec(\theta)\tan(\t heta)\bigg]+C$

    Now remembering that $\displaystyle \theta=artan\bigg(\frac{x}{4}\bigg)$

    we need to imput back in...so seeing that
    $\displaystyle \sec\bigg(arctan\bigg(\frac{x}{4}\bigg)\bigg)=\fra c{\sqrt{x^2+16}}{4}$

    and that $\displaystyle \tan\bigg(arctan\bigg(\frac{x}{4}\bigg)\bigg)=\fra c{x}{4}$

    wee see that

    $\displaystyle \int\sqrt{x^2+16}dx=16\bigg[\frac{1}{2}\bigg[\ln\bigg|\frac{\sqrt{x^2+16}}{4}+\frac{x}{4}\bigg| +\frac{\sqrt{x^2+16}}{4}\cdot\frac{x}{4}\bigg]+C\bigg]=$$\displaystyle 8\bigg[\ln\bigg|\frac{\sqrt{x^2+16}}{4}+\frac{x}{4}\bigg| +\frac{x\sqrt{x^2+16}}{16}\bigg]+C\bigg]$

    Which equals your answer
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