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Math Help - 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)
    \int \sqrt{16+x^2} \,dx

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

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

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

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

    3)
     \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) \int \frac{1}{x^2\sqrt{x-1}}dx

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

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

    no clue
    Did you observe \sqrt{1+sin\,2x} = |\cos x + \sin x| ?
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  3. #3
    Super Member PaulRS's Avatar
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    1.
    Let u=\sin(x)

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

    \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 2\int\frac{du}{(u^2+1)^2}

    Let: z=\arctan(u) and remember that \frac{1}{\tan^2(z)+1}=\cos^2(z)
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  4. #4
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    Did you observe \sqrt{1+sin\,2x} = |\cos x + \sin x| ?
    How does that work?
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  5. #5
    Moo
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    Hello,

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

    \sin 2x=2 \cos x \sin x

    \implies 1+\sin 2x=\cos^2x+\sin^2x+2 \cos x \sin x=(\cos x+\sin x)^2
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  6. #6
    MHF Contributor Mathstud28's Avatar
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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)
    \int \sqrt{16+x^2} \,dx

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

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

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

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

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

    no clue
    For the first one by trig sub we get

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

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

    Imputting this we get

    \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 \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

    \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

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

    and let u=\tan(\theta)

    So v=sec(\theta)

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

    Applying parts we get

    \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

    \int\sec(\theta)+\sec(\theta)\tan^2(\theta)d\theta  =\int\sec^3(\theta)d\theta= \ln|\sec(\theta)+\tan(\theta)|+\sec(\theta)\tan(\t  heta)-\int\sec^3(\theta)d\theta \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

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

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

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

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

    wee see that

    \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]= 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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