Evaluate the following integrals. ∫sin⁡xcos2x+cos⁡xdx

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Evaluate the following integrals.  ∫sin⁡xcos2x+cos⁡xdx

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Serita Dewitt · Accepted Answer

∫sin⁡(x)cos⁡(x)2+cos⁡(x)dx Expression −sin⁡(x) under the sum sign differential, ie.: (−sin⁡(x))dx=d⋅(cos⁡(x)),t=cos⁡(x) Then, starting the integral may be written as: ∫(−1t2+t)dt ∫−1x2+xdx represent a : −1x(x+1)=−1x(x+1) We use the method of decomposition into protozoa. Let us expand the function into the simplest terms: −1x(x+1)=Ax+Bx+1=A(x+1)+Bxx(x+1) Equate the numerators and take into account that the coefficients at the same powers of x on the left and on the right must coincide: -1=A(x+1)+Bx x:A+B=0 1: A=-1 Solving it, we find: A = -1; B = 1; −1x(x+1)=−1x+1x+1 We calculate the tabular integral: We ∫1x+1dx=ln⁡(x+1) calculate the tabular integral: −∫1xdx=−ln⁡(x) Answer: ln⁡(x+1)−ln⁡(x)+C or ln⁡(x+1x)+C To write down the final answer, it remains to substitute cos⁡(x) instead of t. ln⁡(cos⁡(x)+1cos⁡(x))+C

Evaluate the following integrals. \int \frac{\sin x}{\cos^{2}x+\cos x}dx

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