Solve the integral. ∫cos⁡x1+sin2xdx

Question

Solve the integral.   ∫cos⁡x1+sin2xdx

Tiefdruckot · Accepted Answer

Step 1 The given integral is: ∫cos⁡x1+sin2xdx To solve this integral, Consider, sin⁡x=t..(1) Now, differentiate equation (1) on both sides. cos⁡xdx=dt...(2) From equations (1) and (2) plug these values in the given integral. ∫cos⁡x1+sin2xdx=∫(11+t2)dt Step 2 Now, integrate further, ∫cos⁡x1+sin2xdx=tan−1(t)+c Now, from equation (1), again replace t with sin⁡(x). That is, ∫cos⁡x1+sin2xdx=tan−1(sin⁡x)+c ∫cos⁡x1+sin2xdx=tan−1(tan⁡(x1−x2))+c ∫cos⁡x1+sin2xdx=x1−x2+c Where c is an arbitrary constant.

Jack Maxson · Accepted Answer

∫cos⁡(x)1+sin⁡(x)2dx  We put the expression cos⁡(x) under the differential sign, i.e.:  cos⁡(x)dx=d(sin⁡(x)),t=sin⁡(x)  Then the original integral can be written as follows:  ∫1t2+1dt  This is a tabular integral:  ∫1t2+1dt=arctan⁡(t)+C  To write the final The answer is, it remains to substitute sin⁡(x) instead of t.  arctan⁡(sin⁡(x))+C

Vasquez · Accepted Answer

∫cos⁡(x)1+sin⁡(x)2dx∫11+t2dt11×arctan⁡(t1)11×arctan⁡(sin⁡(x)1)arctan⁡(sin⁡(x))Answer:arctan⁡(sin⁡(x))+C

Solve the integral. \int \frac{\cos x}{1+\sin^{2}x}dx

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