If-0-pi-1-a-b-cos-x-dx-a-gt-0-is-equal-to-pi-a-2-b-2-then-0-pi-1-a-b-cos-x-2-dx-

Question Number 64267 by Chi Mes Try last updated on 16/Jul/19

If \underset{0}{\int^{π}} \frac{1}{a + b \cos x} d x, a > 0 is equal to \frac{π}{\sqrt{a^{2} - b^{2}}},

then \underset{0}{\int^{π}} \frac{1}{{(a + b \cos x)}^{2}} d x =

Commented by Prithwish sen last updated on 16/Jul/19

\int_{0}^{π} \frac{d}{da} [\frac{1}{a + bcosx}] dx = \frac{d}{da} \frac{π}{\sqrt{a^{2} - b^{2}}}

\int_{0}^{π} \frac{dx}{{(1 + cosx)}^{2}} = \frac{a π}{{(a^{2} - b^{2})}^{\frac{3}{2}}} please check .

Answered by Tanmay chaudhury last updated on 16/Jul/19

\int_{0}^{π} \frac{d x}{a ({s i n}^{2} \frac{x}{2} + {c o s}^{2} \frac{x}{2}) + b ({c o s}^{2} \frac{x}{2} - {s i n}^{2} \frac{x}{2})}

\int_{0}^{π} \frac{d x}{(a + b) {c o s}^{2} \frac{x}{2} + (a - b) {s i n}^{2} \frac{x}{2}}

\int_{0}^{π} \frac{{s e c}^{2} \frac{x}{2}}{(a + b) + (a - b) {t a n}^{2} \frac{x}{2}}

\frac{1}{a - b} \int_{0}^{π} \frac{{s e c}^{2} \frac{x}{2} d x}{{(\sqrt{\frac{a + b}{a - b}})}^{2} + {t a n}^{2} \frac{x}{2}}

p = t a n \frac{x}{2} \frac{d p}{d x} = {s e c}^{2} \frac{x}{2} \times \frac{1}{2}

\frac{1}{a - b} \int_{0}^{\infty} \frac{2 d p}{{(\sqrt{\frac{a + b}{a - b}})}^{2} + p^{2}}

\frac{2}{a - b} \times \frac{1}{\sqrt{\frac{a + b}{a - b}}} ∣ {t a n}^{- 1} (\frac{p}{\sqrt{\frac{a + b}{a - b}}}) ∣_{0}^{\infty}

= \frac{2}{\sqrt{a^{2} - b^{2}}} \times [{t a n}^{- 1} (\infty)]

= \frac{π}{\sqrt{a^{2} - b^{2}}}

I (a, b) = \int_{0}^{π} \frac{d x}{a + b c o s x} = \frac{π}{\sqrt{a^{2} - b^{2}}}

\frac{d I}{d a} = \int_{0}^{π} \frac{\partial}{\partial a} (\frac{1}{a + b c o s x}) d x = π \frac{\partial}{\partial a} {{(a^{2} - b^{2})}^{\frac{- 1}{2}}}

\frac{d I}{d a} = \int_{0}^{π} \frac{- 1}{{(a + b c o s x)}^{2}} \times (1 + 0) d x = \frac{- π}{2} \times {(a^{2} - b^{2})}^{\frac{- 3}{2}} \times 2 a

s o \int_{0}^{π} \frac{- d x}{{(a + b c o s x)}^{2}} = \frac{- π a}{{(a^{2} - b^{2})}^{\frac{3}{2}}}

s o \int_{0}^{π} \frac{d x}{{(a + b c o s x)}^{2}} = \frac{π a}{{(a^{2} - b^{2})}^{\frac{3}{2}}}

T a n m a y 16.07.19