Ex. 1 Substitution Help (1 Viewer)

bringonthenerds · Oct 3, 2012

1. Use substitution x=2sin(theta)

Integral of root(4-x^2), with restrictions 1 and -1

2. No substitution

Integral of [(sinx)^2 + x ], with restrictions pi to pi/2

Anyone? I'm dying here.

nightweaver066 · Oct 3, 2012

I think you need this:

$cos2x = 1 - 2sin^2x = 2cos^2x - 1$

$$from this we get $ sin^2x = \frac{1}{2}(1 - cos2x)$

$cos^2x = \frac{1}{2}(1 + cos2x)$

For the first one..

$\int^1_{-1} \sqrt{4 - x^2}dx$

$= 2\int^1_0 \sqrt{4 - x^2}dx$

$$Let $ x = 2sin\theta$

$dx = 2cos\theta d\theta$

$x = 1, \theta = \frac{\pi}{6}$

$x = 0, \theta = 0$

$\therefore 2\int^{\frac{\pi}{6}}_0 \sqrt{4 - 4sin^2\theta} \times 2cos\theta d\theta$

$= 2\int^{\frac{\pi}{6}}_0 2cos\theta \times 2cos\theta d\theta$

$= 8\int^{\frac{\pi}{6}}_0 cos^2\theta d\theta$

.. and you can continue on.

Ex. 1 Substitution Help (1 Viewer)

bringonthenerds

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nightweaver066

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