Author Topic: Integration [SOLVED] (Read 417 times) Tweet Share

98.40_for_sure · « **on:** October 04, 2010, 11:44:27 pm »

So like... i'm doing this question, and i can think of two ways to do it, however they should give me the same answer, but they don't...

$\int_{0}^{a}(sin^{2}(\frac{3x}{2})-cos^{2}(\frac{3x}{2}))dx$

My first way is to double angle formula it into $\int_{0}^{a}cos(3x)dx$ where i end up with $\frac{1}{3}sin(3a)$

My second way is to use the trig identity $cos^2(x)+sin^2(x)=1$ where i change either sin or cos and then use the equivalent formula thingie;

$\int_{0}^{a}(sin^{2}(\frac{3x}{2})-cos^{2}(\frac{3x}{2}))dx$

$\int_{0}^{a}2sin^2{(\frac{3x}{2})}-1dx$

$\int_{0}^{a}1-cos(3x)-1dx$

and i end up with the answer $-\frac{1}{3}sin(3a)$

Can you see any mistakes with my working? I've checked it a hundred times and it should work

m@tty · « **Reply #1 on:** October 04, 2010, 11:47:23 pm »

Your first method is wrong.

$\sin^2\left(\frac{3x}{2}\right)-\cos^2\left(\frac{3x}{2}\right)=-\cos(3x)$

98.40_for_sure · « **Reply #2 on:** October 04, 2010, 11:49:42 pm »

*bangs head*
BLOODY CHECKPOINTS IS WRONG... AGAIN. I WAS RIGHT DAMMIT
thanks m@tty

m@tty · « **Reply #3 on:** October 04, 2010, 11:50:44 pm »

No problem :]

So many errors in checkpoints.

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Author Topic: Integration [SOLVED] (Read 417 times) Tweet Share

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Integration [SOLVED]

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