Evaluate the integral. ∫sec2xtan⁡xdx

Question

Evaluate the integral.   ∫sec2xtan⁡xdx

Edward Patten · Accepted Answer

Step 1Given:&amp;nbsp;I=∫sec2xtan⁡x&amp;nbsp;dx&amp;nbsp;we substitute for evaluating a given integraltan⁡x=t...(1)now, differentiating equation (1) with respect to xd&amp;nbsp;dx&amp;nbsp;(tan⁡x)=d&amp;nbsp;dx&amp;nbsp;(t)&amp;nbsp;&amp;nbsp;&amp;nbsp;(∵d&amp;nbsp;dx&amp;nbsp;(tan⁡x)=sec2x)sec2x=&amp;nbsp;dt&amp;nbsp;&amp;nbsp;dx&amp;nbsp;sec2x&amp;nbsp;dx&amp;nbsp;=&amp;nbsp;dt&amp;nbsp;Step 2now, replacing&amp;nbsp;sec2x&amp;nbsp;dx&amp;nbsp;&amp;nbsp;with dt, tanx with t in given integralso,∫sec2xtan⁡x&amp;nbsp;dx&amp;nbsp;=∫t&amp;nbsp;dt&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(∵∫xn&amp;nbsp;dx&amp;nbsp;=xn+1n+1+c)=(t22)+c...(2)now, replacing t with&amp;nbsp;tan⁡x&amp;nbsp;in equation (2)so,∫sec2xtan⁡x&amp;nbsp;dx&amp;nbsp;=tan2x2+cConsequently, the given integral equals&amp;nbsp;tan2x2+c.

Jack Maxson · Accepted Answer

∫sec2(x)tan⁡(x)dx  Substitution u=sec2(x)⇒dudx=2sec2(x)tan⁡(x)⇒dx=12sec2(x)tan⁡(x)du:  =12∫1du  Now we calculate:  ∫1du  Integral of a constant:  =u  We substitute the already calculated integrals:  12∫1du  =u2  Reverse replacement u=sec2(x):  =sec2(x)2  ∫sec2(x)tan⁡(x)dx  =sec2(x)2+C

nick1337 · Accepted Answer

∫sec2⁡(x)tan⁡(x)dx Apply u-substitution: u=tan⁡(x) =∫udu Apply the Power Rule =u22 Substitute back u=tan⁡(x) =tan2⁡(x)2 Add a constant to the solution =tan2⁡(x)2+C

Evaluate the integral. \int \sec^{2}x\tan x dx

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