Evaluate-9-2-0-pi-2-sin-cos-cos-3-sin-3-2-d-

Question Number 2241 by Yozzi last updated on 10/Nov/15

E v a l u a t e

\frac{9}{2} \int_{0}^{π / 2} {(\frac{s i n θ c o s θ}{{c o s}^{3} θ + {s i n}^{3} θ})}^{2} d θ .

Answered by Filup last updated on 11/Nov/15

= \frac{9}{2} \int_{0}^{π / 2} \frac{\sin^{2} θ \cos^{2} θ}{{(\cos^{3} θ + \sin^{3} θ)}^{2}} d θ

= \frac{9}{2} \int_{0}^{π / 2} \frac{\sin^{2} θ \cos^{2} θ}{{(\cos^{3} θ + \sin^{3} θ)}^{2}} \times \frac{\sec^{6} θ}{\sec^{6} θ} d θ

= \frac{9}{2} \int_{0}^{π / 2} \frac{\frac{\sin^{2} θ}{\cos^{4} θ}}{\cos^{6} θ \sec^{6} θ + {2 \cos}^{3} θ \sin^{3} θ \sec^{6} θ + \sin^{6} θ \sec^{6} θ} d θ

= \frac{9}{2} \int_{0}^{π / 2} \frac{\tan^{2} θ \sec^{2} θ}{1 + {2 \sin}^{3} θ \sec^{3} θ + \tan^{6} θ} d θ

= \frac{9}{2} \int_{0}^{π / 2} \frac{\tan^{2} θ \sec^{2} θ}{1 + {2 \tan}^{3} θ + \tan^{6} θ} d θ

= \frac{9}{2} \int_{0}^{π / 2} \frac{\tan^{2} θ \sec^{2} θ}{{(1 + \tan^{3} θ)}^{2}} d θ

u = \tan θ d u = \sec^{2} θ d θ

= \frac{9}{2} \int_{0}^{π / 2} \frac{u^{2} d u}{{(1 + u^{3})}^{2}}

s = u^{3} + 1 d s = 3 u^{2} d u

= \frac{1}{3} \times \frac{9}{2} \int_{0}^{π / 2} \frac{3 u^{2} d u}{{(1 + u^{3})}^{2}}

= \frac{9}{6} \int_{0}^{π / 2} \frac{d s}{s^{2}}

= \frac{3}{2} \int_{0}^{π / 2} \frac{1}{s^{2}} d s

= \frac{3}{2} {[- \frac{1}{s}]}_{0}^{π / 2}

= \frac{3}{2} {[- \frac{1}{u^{3} + 1}]}_{0}^{π / 2}

= \frac{3}{2} {[- \frac{1}{\tan^{3} θ + 1}]}_{0}^{π / 2}

\lim_{x \to {(π / 2)}^{-}} \tan (x) = + \infty

= - \frac{3}{2} (\frac{1}{\lim_{θ \to {(π / 2)}^{-}} (\tan θ) + 1} - \frac{1}{\tan (0) + 1})

= - \frac{3}{2} (\frac{1}{\infty} - \frac{1}{1}) = - \frac{3}{2} (- 1)

= \frac{3}{2}

∴ \frac{9}{2} \int_{0}^{π / 2} {(\frac{\sin θ \cos θ}{\cos^{3} θ + \sin^{3} θ})}^{2} d θ = \frac{3}{2}

Commented by Filup last updated on 11/Nov/15

Note :

\lim_{x \to {(π / 2)}^{+}} \tan (x) = - \infty

\lim_{x \to {(π / 2)}^{-}} \tan (x) = + \infty

The reasoning for my chioce of \tan (π / 2)

is that for the intgral, x \in [0, π / 2] .

Thus as x \to π / 2, where x \in [0, π / 2],

f (x) \to + \infty .

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