Trig Substitution Integral

**joatmon** · August 30th 2011, 09:29 PM

$\displaystyle \int_0^\frac{\pi}{2} \frac{\cos t dt}{\sqrt{1 + \sin ^2 t}}$

Here's what I have done:

$let x = \sin t \rightarrow dx = \cos t dt , x(\frac{\pi}{2}) = 1, x(0) = 0$

$\int_0^1 \frac {dx}{\sqrt{1+x^2}}$

Visualizing a right triangle with the horizontal side = 1 and the vertical side = x,

$\sec\theta = \sqrt{1+x^2}$
$\tan\theta = x \rightarrow dx = \sec^2\theta d\theta$

$x = 1 \rightarrow \theta = \frac{\pi}{4} \rightarrow \sec\theta = \sqrt{2}$

$x = 0 \rightarrow \theta = 0 \rightarrow \sec\theta = 1$

$\int_1^{\sqrt{2}} \frac{\sec^2 \theta d\theta}{\sec\theta}$

$\int_1^{\sqrt{2}} \sec\theta d\theta$

$\ln \mid \sec\theta + \tan\theta \mid_1^\sqrt{2}$

$\ln (\sec\sqrt{2} + \tan\sqrt{2}) - \ln(\sec(1) + \tan(1))$

I thought that I had this, but the answer doesn't check out. Can anybody see what I am missing? Thanks again!

**Prove It** · August 30th 2011, 09:35 PM

Originally Posted by joatmon

$\displaystyle \int_0^\frac{\pi}{2} \frac{\cos t dt}{\sqrt{1 + \sin ^2 t}}$

Here's what I have done:

$let x = \sin t \rightarrow dx = \cos t dt , x(\frac{\pi}{2}) = 1, x(0) = 0$

$\int_0^1 \frac {dx}{\sqrt{1+x^2}}$

Visualizing a right triangle with the horizontal side = 1 and the vertical side = x,

$\sec\theta = \sqrt{1+x^2}$
$\tan\theta = x \rightarrow dx = \sec^2\theta d\theta$

$x = 1 \rightarrow \theta = \frac{\pi}{4} \rightarrow \sec\theta = \sqrt{2}$

$x = 0 \rightarrow \theta = 0 \rightarrow \sec\theta = 1$

$\int_1^{\sqrt{2}} \frac{\sec^2 \theta d\theta}{\sec\theta}$

$\int_1^{\sqrt{2}} \sec\theta d\theta$

$\ln \mid \sec\theta + \tan\theta \mid_1^\sqrt{2}$

$\ln (\sec\sqrt{2} + \tan\sqrt{2}) - \ln(\sec(1) + \tan(1))$

I thought that I had this, but the answer doesn't check out. Can anybody see what I am missing? Thanks again!

When you make the substitution $\displaystyle x = \tan{\theta} \implies dx = \sec^2{\theta}\,d\theta$ , you are converting the integrand to a function of $\displaystyle \theta$ , which means your terminals need to be values of $\displaystyle \theta$ , not $\displaystyle \sec{\theta}$ .

So it should be $\displaystyle \int_0^{\frac{\pi}{4}}{\frac{\sec^2{\theta}}{\sqrt {1 + \tan^2{\theta}}}\,d\theta}$ .

But on an entirely different note, I believe hyperbolic substitution will be easier than trigonometric in this case.

Let $\displaystyle x = \sinh{t} \implies dx = \cosh{t}\,dt$ .

**joatmon** · August 30th 2011, 10:04 PM

Thanks!

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