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# Integrate 1/(ax+a^2-x^2) from 0 to a and sin^7 x from -pi/2 to pi/2

Jitender Singh IIT Delhi
7 years ago
Ans:
2.
$I = \int_{-\pi /2}^{\pi /2}sin^{7}(x)dx$
$I = \int_{-a}^{a}f(x)dx = \int_{0}^{a}[f(x)+f(-x)]dx$
$I = \int_{0}^{\pi /2}(sin^{7}(x)+sin^{7}(-x))dx$
$I = \int_{0}^{\pi /2}(sin^{7}(x)-sin^{7}(x))dx = 0$
1.
$I = \int_{0}^{a}\frac{1}{ax+a^{2}-x^{2}}dx$
$I = \int_{0}^{a}\frac{1}{ax+\frac{5a^{2}}{4}-\frac{a^{2}}{4}-x^{2}}dx$
$I = \int_{0}^{a}\frac{1}{\frac{5a^{2}}{4}-(\frac{a^{2}}{4}+x^{2}-ax)}dx$
$I = \int_{0}^{a}\frac{1}{\frac{5a^{2}}{4}-(x-\frac{a}{2})^{2}}dx$
$I = \int_{0}^{a}\frac{1}{(\frac{\sqrt{5}a}{2})^{2}-(x-\frac{a}{2})^{2}}dx$
$I = \frac{2}{2\sqrt{5}a}(ln(\frac{\frac{\sqrt{5}a}{2}+x-\frac{a}{2}}{\frac{\sqrt{5}a}{2}-x+\frac{a}{2}}))_{0}^{a}$
$I = \frac{1}{\sqrt{5}a}(ln(\frac{\sqrt{5}+1}{\sqrt{5}-1})-ln(\frac{\sqrt{5}-1}{\sqrt{5}+1}))$
$I = \frac{2}{\sqrt{5}a}ln(\frac{\sqrt{5}+1}{\sqrt{5}-1})$
Thanks & Regards
Jitender Singh
IIT Delhi