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Math Help - Integrating Cosine and sine integrals by substitution

  1. #1
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    Integrating Cosine and sine integrals by substitution

    Hi all,

    I have a few integrals that I have been asked to evaluate using integration by substitution. I have never came across anything like this before.

    For example;

    Integrate (1+sinx)/cosx by integration.

    Is it possible. I know how to integrate other forms of integrals by substitution for example involving x's.

    Thanks.
    Last edited by Googl; June 28th 2011 at 02:56 AM.
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    Re: Integrating Cosine and sine integrals by substitution

    Looks like I am getting some where.

    Let u = 1+sinx, du/dx = cosx, du = cosxdx

    I am supposed to write everything in terms of u. But to replace cosxdx the other cosx to replace is a fraction of 1/cosx
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    Re: Integrating Cosine and sine integrals by substitution

    Quote Originally Posted by Googl View Post
    Hi all,

    I have a few integrals that I have been asked to evaluate using integration by substitution. I have never came across anything like this before.

    For example;

    Integrate (1+sinx)/cosx by integration.

    Is it possible. I know why integrate other forms of integrals by substitution for example involving x's.

    Thanks.
    Dear Googl,

    Substitute, \sin x=u
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    Re: Integrating Cosine and sine integrals by substitution

    Spoiler:
    \begin{aligned}\int \frac{1+\sin{x}}{\cos{x}}\;{dx} = -\int \frac{\cos \left (\frac{x}{2} - \frac{\pi}{4} \right )}{\sin \left ( \frac{x}{2} - \frac{\pi}{4} \right )}\;{dx} = -2\int \frac{\left[\sin \left (\frac{x}{2} - \frac{\pi}{4} \right )\right]'}{\sin \left ( \frac{x}{2} - \frac{\pi}{4} \right )}\;{dx} = \ln\bigg| \csc^2\left ( \frac{x}{2} - \frac{\pi}{4} \right )\bigg|+\mathcal{C}.\end{aligned}
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    Re: Integrating Cosine and sine integrals by substitution

    Quote Originally Posted by Googl View Post
    Hi all,

    I have a few integrals that I have been asked to evaluate using integration by substitution. I have never came across anything like this before.

    For example;

    Integrate (1+sinx)/cosx by integration.

    Is it possible. I know why integrate other forms of integrals by substitution for example involving x's.

    Thanks.
    \displaystyle \begin{align*} \int{\frac{1 + \sin{x}}{\cos{x}}\,dx} &= \int{\frac{(1 + \sin{x})(1 - \sin{x})}{\cos{x}(1 - \sin{x})}\,dx} \\ &= \int{\frac{1 - \sin^2{x}}{\cos{x}(1 - \sin{x})}\,dx} \\ &= \int{\frac{\cos^2{x}}{\cos{x}(1 - \sin{x})}\,dx} \\ &= \int{\frac{\cos{x}}{1 - \sin{x}}\,dx} \\ &= -\int{\frac{-\cos{x}}{1 - \sin{x}}\,dx} \\ &= -\int{\frac{1}{u}\,du}\textrm{ after making the substitution }u = 1 - \sin{x} \implies du = -\cos{x}\,dx\end{align*}

    I'm sure you can go from here...
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    Re: Integrating Cosine and sine integrals by substitution

    Quote Originally Posted by Sudharaka View Post
    Dear Googl,

    Substitute, \sin x=u
    Could you continue a bit further. The problem is that du results in cosxdx.
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    Re: Integrating Cosine and sine integrals by substitution

    Quote Originally Posted by Prove It View Post
    \displaystyle \begin{align*} \int{\frac{1 + \sin{x}}{\cos{x}}\,dx} &= \int{\frac{(1 + \sin{x})(1 - \sin{x})}{\cos{x}(1 - \sin{x})}\,dx} \\ &= \int{\frac{1 - \sin^2{x}}{\cos{x}(1 - \sin{x})}\,dx} \\ &= \int{\frac{\cos^2{x}}{\cos{x}(1 - \sin{x})}\,dx} \\ &= \int{\frac{\cos{x}}{1 - \sin{x}}\,dx} \\ &= -\int{\frac{-\cos{x}}{1 - \sin{x}}\,dx} \\ &= -\int{\frac{1}{u}\,du}\textrm{ after making the substitution }u = 1 - \sin{x} \implies du = -\cos{x}\,dx\end{align*}

    I'm sure you can go from here...
    Why did you multiply by (1 - sinx) at the beginning?
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    Re: Integrating Cosine and sine integrals by substitution

    Quote Originally Posted by Googl View Post
    Could you continue a bit further. The problem is that du results in cosxdx.
    Dear Googl,

    \sin x=u\Rightarrow \cos x=\sqrt{1-u^2}

    du=\cos x~dx=\sqrt{1-u^2}~dx\Rightarrow dx=\frac{du}{\sqrt{1-u^2}}

    {\int\frac{1+sin x}{\cos x}=\int\frac{1+u}{\sqrt{1-u^2}}\frac{du}{\sqrt{1-u^2}}=\int\frac{du}{1-u}=-\ln|u-1|+C=-\ln|\sin x-1|+C}

    Hope this will help you.
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    Re: Integrating Cosine and sine integrals by substitution

    Quote Originally Posted by Googl View Post
    Looks like I am getting some where.

    Let u = 1+sinx, du/dx = cosx, du = cosxdx

    I am supposed to write everything in terms of u. But to replace cosxdx the other cosx to replace is a fraction of 1/cosx
    Your substitution can also be used. Try to write cos x in terms of u and substitute it as I had done in my previous post. Hope you can continue.
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    Re: Integrating Cosine and sine integrals by substitution

    Quote Originally Posted by Googl View Post
    Why did you multiply by (1 - sinx) at the beginning?
    Because it's easier to make sure you have a single sine or cosine in the numerator, so that it can become part of your \displaystyle du...
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    Re: Integrating Cosine and sine integrals by substitution

    I will think through your examples and get back to you.

    Thanks.
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    Re: Integrating Cosine and sine integrals by substitution

    straight-forward integration ...

    \int \frac{1+\sin{x}}{\cos{x}} \, dx =

    \int \sec{x} + \tan{x} \, dx =

    \ln|\sec{x}+\tan{x}| - \ln|\cos{x}| + C = \ln \left|\frac{1}{1-\sin{x}} \right| + C
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