Evaulating the trigonometric integral ∫1(x2+1)2dxTo do this, I let x=tan⁡u. Now we have dx=sec2udu∫1(x2+1)2dx=∫sec2{u}du(tan2{u}+1)2∫1(x2+1)2dx=∫1sec2{u}du=∫cos2{u}du∫1(x2+1)2dx=∫cos⁡{(2u)}+12du=sin⁡(u)4+u2∫1(x2+1)2dx=1−cos2{u}4+u2Now, I think I am right so far but I do not know have to get rid of the u in the cos2(u) term. Please help.

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haiguetenteme7zyu · Accepted Answer

Since you have x=tan⁡u , think of u as an angle. Thentan⁡u=x1Draw a right triangle with legs x and 1 to demonstrate this fact (with u as the angle opposite the x.) The hypotenuse is 1+x2 Now you can evaluate any trig function of u that you please. E.g.,cos⁡u=11+x2So to evaluatesin⁡2u=2sin⁡ucos⁡u=2x1+x211+x2=2x1+x2

LiaisyAciskriuu · Accepted Answer

∫1(x2+1)2dxApply Trig Substitution: x=tan⁡(u)Then differentiate with respect to &#039;x&#039; on both sides, then we get  dx=sec2(u)du=∫1cos2(u)(tan2(u)+1)2duUsing the identity: tan2(x)=−1+sec2(x)=∫1(1−1+sec2(u))2cos2(u)duNotice that1(1−1+sec2(u))2cos2(u)=1sec4(u)cos2(u)=∫1sec4(u)cos2(u)du=∫cos2(u)duUsing the identity: cos2(x)=1+cos⁡(2x)2=∫1+cos⁡(2u)2du=12∫1+cos⁡(2u)du=12∫1du+∫cos⁡(2u)du=12u+12sin⁡(2u)After substituting back u=arctan⁡(x)∫1(x2+1)2dx=12arctan⁡(x)+12sin⁡(2arctan⁡(x))+C

Evaulating the trigonometric integral \(\displaystyle\int{\frac{{{1}}}{{{\left({x}^{{2}}+{1}\right)}^{{2}}}}}{\left.{d}{x}\right.}\) To do this,

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