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Question Number 88603 by hovero clinton last updated on 11/Apr/20

	Answered by TANMAY PANACEA. last updated on 11/Apr/20
	$x=sina ∫_0 ^(π/2) ((lnsina×cosada)/((1+8sin^2 a)cosa)) ∫_0 ^(π/2) lnsina.(1/({1+4(1−cos2a)})) ∫_0 ^(π/2) lnsina.(1/({5−4.((1−tan^2 a)/(1+tan^2 a))}))da ∫_0 ^(π/2) lnsina.((sec^2 a)/((5+5tan^2 a−4+4tan^2 a)))da ∫_0 ^(π/2) lnsina.((sec^2 a)/(1+9tan^2 a))da now lnsina∫((sec^2 a da)/(1+9tan^2 a))−∫[cota∫((sec^2 a)/(1+9tan^2 a))da]da lnsina∫((d(tana))/(9((1/9)+tan^2 a)))−∫cota.∫((d(tana))/(9((1/9)+tan^2 a)))da lnsina×(1/9)×3×tan^(−1) (((tana)/(1/3)))−∫cota.(1/9).3tan^(−1) (((tana)/(1/3)))da ((lnsina.tan^(−1) (3tana))/3)−(1/3)∫cota.tan^(−1) (3tana) wait...$
	$x = s i n a$ $\int_{0}^{\frac{π}{2}} \frac{l n s i n a \times c o s a d a}{(1 + 8 {s i n}^{2} a) c o s a}$ $\int_{0}^{\frac{π}{2}} l n s i n a . \frac{1}{{1 + 4 (1 - c o s 2 a)}}$ $\int_{0}^{\frac{π}{2}} l n s i n a . \frac{1}{{5 - 4 . \frac{1 - {t a n}^{2} a}{1 + {t a n}^{2} a}}} d a$ $\int_{0}^{\frac{π}{2}} l n s i n a . \frac{{s e c}^{2} a}{(5 + 5 {t a n}^{2} a - 4 + 4 {t a n}^{2} a)} d a$ $\int_{0}^{\frac{π}{2}} l n s i n a . \frac{{s e c}^{2} a}{1 + 9 {t a n}^{2} a} d a$ $n o w$ $l n s i n a \int \frac{{s e c}^{2} a d a}{1 + 9 {t a n}^{2} a} - \int [c o t a \int \frac{{s e c}^{2} a}{1 + 9 {t a n}^{2} a} d a] d a$ $l n s i n a \int \frac{d (t a n a)}{9 (\frac{1}{9} + {t a n}^{2} a)} - \int c o t a . \int \frac{d (t a n a)}{9 (\frac{1}{9} + {t a n}^{2} a)} d a$ $l n s i n a \times \frac{1}{9} \times 3 \times {t a n}^{- 1} (\frac{t a n a}{\frac{1}{3}}) - \int c o t a . \frac{1}{9} . 3 {t a n}^{- 1} (\frac{t a n a}{\frac{1}{3}}) d a$ $\frac{l n s i n a . {t a n}^{- 1} (3 t a n a)}{3} - \frac{1}{3} \int c o t a . {t a n}^{- 1} (3 t a n a)$ $w a i t . . .$
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