solve the following integral 1)∫_3 ^7 4(√((x−3)(7−x)))dx 2)∫_0 ^∞ ((xln^2 (1+x))/((1+x)^3 ))dx 3)∫_0 ^(π/4) [ln(1−tanx)]^2 dx=(π/2)ln2−2G 4)∫_0 ^(π/4) ln(1+cotx)dx=(π/8)ln2+G 5)∫


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Question Number 109734 by mathdave last updated on 25/Aug/20

$s o l v e t h e f o l l o w i n g i n t e g r a l$ $1) \int_{3}^{7} 4 \sqrt{(x - 3) (7 - x)} d x$ $2) \int_{0}^{\infty} \frac{x \ln^{2} (1 + x)}{{(1 + x)}^{3}} d x$ $3) \int_{0}^{\frac{π}{4}} {[\ln (1 - \tan x)]}^{2} d x = \frac{π}{2} \ln 2 - 2 G$ $4) \int_{0}^{\frac{π}{4}} \ln (1 + \cot x) d x = \frac{π}{8} \ln 2 + G$ $5) \int_{0}^{\frac{π}{2}} \ln (2 + \cos x) d x$
	Answered by bobhans last updated on 25/Aug/20
	$(1) ∫_3 ^7 4(√(4−(5−x)^2 )) dx = I let 5−x=2sin w → { ((w=π/2)),((w=−π/2)) :} I=∫_(π/2) ^(−π/2) 4(√(4−4sin^2 w)) (−2cos w dw) I= ∫_(−π/2) ^(π/2) 16 cos^2 w dw = 16∫_(−π/2) ^(π/2) ((1/2)+(1/2)cos 2w)dw =8[w+(1/2)sin 2w]_(−π/2) ^(π/2) = 8[π+0]=8π$
	$(1) \underset{3}{\int^{7}} 4 \sqrt{4 - {(5 - x)}^{2}} d x = I$ $l e t 5 - x = 2 \sin w \to {\begin{cases} w = π / 2 \\ w = - π / 2 \end{cases}$ $I = \underset{π / 2}{\int^{- π / 2}} 4 \sqrt{4 - 4 \sin^{2} w} (- 2 \cos w d w)$ $I = \underset{- π / 2}{\int^{π / 2}} 16 \cos^{2} w d w = 16 \underset{- π / 2}{\int^{π / 2}} (\frac{1}{2} + \frac{1}{2} \cos 2 w) d w$ $= 8 {[w + \frac{1}{2} \sin 2 w]}_{- π / 2}^{π / 2} = 8 [π + 0] = 8 π$
	Commented by mathdave last updated on 25/Aug/20

	$t h a t 4 n e a r t h e r o o t i s f o u r t h r o o t o o o o$
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