Evaluate the integral. tan⁡xsec3xdx

Question

Evaluate the integral.  tan⁡xsec3xdx

Alex Sheppard · Accepted Answer

Step 1GivenEvaluate the integral.∫tan⁡xsec3xdxStep 2Let u=sec⁡(x).Thendu=(sec⁡(x))′dx=tan⁡(x)sec⁡(x)dxdu=(sec⁡(x))′dx=tan⁡(x)sec⁡(x)dx,and we have thattan⁡(x)sec⁡(x)dx=du.Therefore,∫tan⁡(x)sec3(x)dx=∫u2duApply the power rule ∫undu=un+1n+1(n≠−1)∫u2du=u1+21+2=(u33)Recall that u=sec⁡(x)u=sec⁡(x):u33=sec⁡(x)33Therefore,∫tan⁡(x)sec3(x)dx=sec3(x)3Add the constant of integration:∫tan⁡(x)sec3(x)dx=sec3(x)3+C

Pademagk71 · Accepted Answer

∫tan⁡xsec3⁡xdx=∫sec2⁡xtan⁡xsec⁡xdx=[u=sec⁡xdu=tan⁡xsec⁡xdx]  =∫u2du=u33+C=sec3x3+C  Result:  sec3x3+C

karton · Accepted Answer

Step 1Apply u-substitutionLet u=sec3⁡(x)⇒dx=13sec3⁡(x)tan⁡(x)Thus,∫tan⁡(x)sec3⁡(x)dx=13∫1du=13u+C=sec3⁡(x)3+CResult:sec3⁡(x)3+C

Evaluate the integral. tan⁡xsec3xdx

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