0-pi-2-1-4sin-2-x-5cos-2-x-dx-

Question Number 83781 by john santu last updated on 06/Mar/20

\int_{0}^{\frac{π}{2}} \frac{1}{4 \sin^{2} x + 5 \cos^{2} x} d x

Commented by niroj last updated on 06/Mar/20

\int_{0}^{\frac{π}{2}} \frac{1}{{4 \sin}^{2} x + {5 \cos}^{2} x} dx

= \int_{0}^{\frac{π}{2}} \frac{\sec^{2} x dx}{{4 \tan}^{2} x + 5}

put \tan x = t

\sec^{2} xdx = dt

if x = \frac{π}{2} then t = \infty

if x = 0 then t = 0

= \int_{0}^{\infty} \frac{1}{{4 t}^{2} + 5} dt

= \frac{1}{4} \int_{0}^{\infty} \frac{1}{t^{2} + \frac{5}{4}} dt

= \frac{1}{4} {[\int \frac{1}{{(t)}^{2} + {(\frac{\sqrt{5}}{2})}^{2}} dt]}_{0}^{\infty}

= \frac{1}{4} {[\frac{1}{\frac{\sqrt{5}}{2}} \tan^{- 1} \frac{t}{\frac{\sqrt{5}}{2}}]}_{0}^{\infty}

= \frac{1}{4} [\frac{2}{\sqrt{5}} \tan^{- 1} (\infty) . \frac{2}{\sqrt{5}} - 0]

= \frac{1}{4} . \frac{2}{\sqrt{5}} . \frac{π}{2} = \frac{π}{4 \sqrt{5}} / / .

Commented by john santu last updated on 06/Mar/20

in short cut

= \int_{0}^{\frac{π}{2}} \frac{dx}{a^{2} \sin^{2} x + b^{2} \cos^{2} x}

= \frac{π}{2 ab} [a = \sqrt{4} = 2, b = \sqrt{5}]

= \frac{π}{2.2 . \sqrt{5}} = \frac{π}{4 \sqrt{5}} ★

Answered by MJS last updated on 06/Mar/20

\int \frac{d x}{{4 \sin}^{2} x + {5 \cos}^{2} x} =

[t = \tan x \to d x = \frac{d t}{t^{2} + 1}; \sin x = \frac{t}{\sqrt{t^{2} + 1}}; \cos x = \frac{1}{\sqrt{t^{2} + 1}}]

= \int \frac{d t}{4 t^{2} + 5} = \frac{\sqrt{5}}{10} \arctan \frac{2 \sqrt{5} t}{5} =

= \frac{\sqrt{5}}{10} \arctan \frac{2 \sqrt{5} \tan x}{5} + C

snswer is \frac{π \sqrt{5}}{20}

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