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Question Number 149805 by mnjuly1970 last updated on 07/Aug/21

	Answered by Ar Brandon last updated on 07/Aug/21
	$I=∫_0 ^(π/2) csc^2 xln(1+2sin^2 x)dx { ((u(x)=ln(1+2sin^2 x))),((v′(x)=csc^2 x)) :}⇒ { ((u′(x)=((2sin2x)/(1+2sin^2 x)))),((v(x)=−cotx)) :} I=[−cotx∙ln(1+2sin^2 x)]_0 ^(π/2) +4∫_0 ^(π/2) ((sinxcos^2 x)/(sinx+2sin^3 x))dx =−4∫_0 ^(π/2) ((sin^2 x−1)/(2sin^2 x+1))dx=−2∫_0 ^(π/2) (1−(3/(2sin^2 x+1)))dx =−π+6∫_0 ^(π/2) ((sec^2 x)/(2tan^2 x+sec^2 x))dx=−π+6∫_0 ^(π/2) ((d(tanx))/(3tan^2 x+1)) =−π+2∙(√3)[arctan((√3)tanx)]_0 ^(π/2) =π((√3)−1)$
	$I = \int_{0}^{\frac{π}{2}} \csc^{2} x \ln (1 + {2 \sin}^{2} x) d x$ ${\begin{cases} u (x) = \ln (1 + {2 \sin}^{2} x) \\ v^{'} (x) = \csc^{2} x \end{cases} \Rightarrow {\begin{cases} u^{'} (x) = \frac{2 \sin 2 x}{1 + {2 \sin}^{2} x} \\ v (x) = - \cot x \end{cases}$ $I = {[- \cot x \cdot \ln (1 + {2 \sin}^{2} x)]}_{0}^{\frac{π}{2}} + 4 \int_{0}^{\frac{π}{2}} \frac{\sin x \cos^{2} x}{\sin x + {2 \sin}^{3} x} d x$ $= - 4 \int_{0}^{\frac{π}{2}} \frac{\sin^{2} x - 1}{{2 \sin}^{2} x + 1} d x = - 2 \int_{0}^{\frac{π}{2}} (1 - \frac{3}{{2 \sin}^{2} x + 1}) d x$ $= - π + 6 \int_{0}^{\frac{π}{2}} \frac{\sec^{2} x}{{2 \tan}^{2} x + \sec^{2} x} d x = - π + 6 \int_{0}^{\frac{π}{2}} \frac{d (\tan x)}{{3 \tan}^{2} x + 1}$ $= - π + 2 \cdot \sqrt{3} {[\arctan (\sqrt{3} \tan x)]}_{0}^{\frac{π}{2}} = π (\sqrt{3} - 1)$
	Commented by mnjuly1970 last updated on 07/Aug/21

	$t h x a l o t m a s t e r . . . (m r b r a n d o n)$
	Commented by Ar Brandon last updated on 07/Aug/21

	${You}^{'} re welcome!$ $How did you celebrate your birthday, Sir ?$ $I guess july 1970 represents your month and$ $year of birth .$ 😄😅
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