# Integrating Cosine and sine integrals by substitution

• Jun 28th 2011, 01:31 AM
Googl
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.
• Jun 28th 2011, 01:40 AM
Googl
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
• Jun 28th 2011, 01:47 AM
Sudharaka
Re: Integrating Cosine and sine integrals by substitution
Quote:

Originally Posted by Googl
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, $\displaystyle \sin x=u$
• Jun 28th 2011, 01:58 AM
TheCoffeeMachine
Re: Integrating Cosine and sine integrals by substitution
Spoiler:
\displaystyle \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}
• Jun 28th 2011, 01:59 AM
Prove It
Re: Integrating Cosine and sine integrals by substitution
Quote:

Originally Posted by Googl
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 \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...
• Jun 28th 2011, 01:59 AM
Googl
Re: Integrating Cosine and sine integrals by substitution
Quote:

Originally Posted by Sudharaka
Dear Googl,

Substitute, $\displaystyle \sin x=u$

Could you continue a bit further. The problem is that du results in cosxdx.
• Jun 28th 2011, 02:04 AM
Googl
Re: Integrating Cosine and sine integrals by substitution
Quote:

Originally Posted by Prove It
\displaystyle \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?
• Jun 28th 2011, 02:13 AM
Sudharaka
Re: Integrating Cosine and sine integrals by substitution
Quote:

Originally Posted by Googl
Could you continue a bit further. The problem is that du results in cosxdx.

Dear Googl,

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

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

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

• Jun 28th 2011, 02:24 AM
Sudharaka
Re: Integrating Cosine and sine integrals by substitution
Quote:

Originally Posted by Googl
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.
• Jun 28th 2011, 02:41 AM
Prove It
Re: Integrating Cosine and sine integrals by substitution
Quote:

Originally Posted by Googl
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 $\displaystyle \displaystyle du$...
• Jun 28th 2011, 05:44 PM
Googl
Re: Integrating Cosine and sine integrals by substitution
I will think through your examples and get back to you.

Thanks.
• Jun 28th 2011, 06:34 PM
skeeter
Re: Integrating Cosine and sine integrals by substitution
straight-forward integration ...

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

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

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