For a minor class i am taking this year, I found the following integral in a problem set, and where i had no luck in evaluating it: ∫cos⁡(2cot−11−x1+x)dx I proceeded as follows: -&gt;first, let x=cos⁡(2θ)⇒dx=−2sin⁡2θdθ so the integral becomes: ∫cos⁡(2cot−11−cos⁡(2θ)1+cos⁡(2θ))−2sin⁡2θdθ=∫cos⁡(2cot−1sin2θcos2θ)−2sin⁡2θdθ =∫cos⁡(2cot−1(tan⁡θ)−2sin⁡2θdθ)=∫cos⁡2θ−2sin⁡2θdθ After this i am stuck. How do i proceed?

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For a minor class i am taking this year, I found the following integral in a problem set, and where i had no luck in evaluating it:  ∫cos⁡(2cot−11−x1+x)dx  I proceeded as follows:  -&amp;gt;first, let x=cos⁡(2θ)⇒dx=−2sin⁡2θdθ so the integral becomes:  ∫cos⁡(2cot−11−cos⁡(2θ)1+cos⁡(2θ))−2sin⁡2θdθ=∫cos⁡(2cot−1sin2θcos2θ)−2sin⁡2θdθ  =∫cos⁡(2cot−1(tan⁡θ)−2sin⁡2θdθ)=∫cos⁡2θ−2sin⁡2θdθ  After this i am stuck. How do i proceed?

boronganfh · Accepted Answer

You can proceed by noting that tan⁡θ=cot⁡(π2−θ) (if you so wish)  Then:  ∫cos⁡(2cot−1(tan⁡θ))⋅−2sin⁡2θdθ  =∫cos⁡(π−2θ)⋅−2sin⁡2θdθ   =∫−cos⁡(2θ)⋅−2sin⁡2θdθ   =∫−xdx   but that is equivalent to the identity

RizerMix · Accepted Answer

Remember cosine double angle identity tells us that cos⁡2θ=cos2⁡θ−sin2⁡θ Here θ=tan−1⁡1+x1−x Drawing a triangle, this means the hypotenuse has to be (1+x)2+(1−x)∗2=2 which means we have that cos⁡θ=1−x2 sin⁡θ=1+x2 cos⁡2θ=1−x2−1+x2=−x and the integral is simply ∫−xdx=−12x2+C

alenahelenash · Accepted Answer

cos⁡2θ=2cos2⁡θ−1 (and there are other choices for the double-angle formula for cosine). So cos⁡(2cot−1⁡1−x1+x)=2cos2⁡(cot−1⁡1−x1+x)−1=2(1−x2)2−1=−x where we use the fact that the cotangent is the adjacent over the opposite (with hypotenuse (1+x)2+(1−x)2=2) and the cosine is the adjacent over the hypotenuse, so is 1−x2

For a minor class i am taking this year, I found the following integral in a problem set, and where

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