Trigonometric integrals Evaluate the following integrals. ∫tan2xdx

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Trigonometric integrals Evaluate the following integrals.  ∫tan2xdx

zesponderyd · Accepted Answer

Step 1  Explanation:  Given that,   ∫tan2xdx  We know that,   tan2x=sec2x−1  Now integrate   ∫tan2xdx=∫sec2x−1dx  ∫tan2xdx=∫sec2xdx−∫1dx  ∫tan2xdx=tan⁡x−x+C  Step 2  Hence,   The integral of ∫tan2xdx is tan⁡x−x+C where C is integral constant.

Jim Hunt · Accepted Answer

∫tan2xdx  Expand the expression  ∫sec⁡(x)2−1dx  ∫sec⁡(x)2dx−∫1dx  Evaluate  tan⁡(x)−x  Add C  Final Answer:  tan⁡(x)−x+C

Vasquez · Accepted Answer

∫tan⁡(x)2dx We make a trigonometric substitution: tan⁡(x)= and then dt=1/(1=t2) ∫t2t2+1dt Simplify the expression: The ∫x2x2+1dx degree of the numerator P(x) is greater than or equal to the degree of the denominator Q(x), so we divide the polynomials. x2x2+1=1+−1x2+1 Integrating the integer part, we get: ∫1dx=x Integrating further, we get: ∫−1x2+1dx=−arctan⁡(x) Answer: x−arctan⁡(x)+C Returning to the change of variables (t=tan⁡(x)), we get: I=tan⁡(x)−arctan⁡(tan⁡(x))+C

Trigonometric integrals Evaluate the following integrals. \int \tan^{2}xdx

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