integration no.7
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Originally Posted by afeasfaerw23231233 integration no.7 $\displaystyle \int \frac{1 - 4u^2}{u^2 + 1} du$ $\displaystyle \int \frac{1}{u^2 + 1}du - 4 \int \frac{u^2}{u^2 + 1} du$ Does this help? -Dan
i know how to do it now , thanks!
The denominator (1 +sin^2(theta)) is not (1 +u^2) because u = cos(theta). 1 +sin^2(theta) = 1 +1 -cos^2(theta) = 2 -cos^2(theta) = 2 -u^2 So it must be INT.[(1 -4u^2) / (2 -u^2)]du
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