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Question

# ∫sin−113(x).cos−13(x)dx

A

3.cot23(x)73.cot83(x)5+c

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B

3.cot23(x)23.cot83(x)5+c

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C

3.cot23(x)23.cot83(x)8+c

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D

3.cot23(x)53.cot83(x)8+c

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Solution

## The correct option is C −3.cot23(x)2−3.cot83(x)8+c We can see that the given integral is also ∫sinmx.cosnx dx . Where m & n are rational numbers. So if m+n−22 is negative then we can either substitute cotx or tanx = t whichever is found suitable. Here, m+n−22=(−113)+(−13)−22=−3 So, we can either substitute cotx = t or tanx = t ∫sin−113(x).cos−13(x)dx Let’s rewrite the expression to get either tanx or cotx in it ∫cos−13(x)sin−13(x).sin4(x)dxOr ∫cot−13(x)sin4(x)dxOr ∫cot−13(x)(cosec2x)2dxOr ∫cot−13(x)(1+cot2(x))(cosec2(x))dx Let’s substitute cot (x) = t −cosec2x.dx=dt−∫t−13.(1+t2)dt=−∫t−13.dt−∫t53.dt−t2323−t8383+c Let’s substitute t = cot(x) in the above expression - −3.cot23(x)2−3.cot83(x)8+c

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