Question

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Answer 1

$\bold{Answer :}$

Now,

∫ (cotx + x) cot²x dx

= ∫ (cotx + x) (cosec²x - 1) dx [∵cosec²x - cot²x = 1]

= ∫ cotx cosec²x dx - ∫ cotx dx + ∫ x cosec²x dx - ∫ x dx ...(1)
______

∴ ∫ cotx cosec²x dx

= ∫ cotx d (- cotx)

= - (cot²x)/2

Here, ∫ cotx dx

= log (sinx)

∴ ∫ x cosec²x dx

= x ∫ cosec²x dx - ∫ {d/dx (x) × ∫ cosec²x dx} dx

= x (- cotx) - ∫ (- cotx) dx [∵ d/dx (x) = 1 ]

= - x cotx + ∫ cotx dx

= - x cotx + log (sinx)

∴ ∫ x dx

= (x²)/2

From (i), we get the required integral as

∫ (cotx + x) cot²x dx

= - (cot²x)/2 - log (sinx) - x cotx + log (sinx) - (x²)/2 + c, where c is integral constant

= - (cot²x)/2 - x cotx - (x²)/2 + c

#$\bold{MarkAsBrainliest}$

Answer 2

Answer:

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