Evaluate the integral. ∫dxcos⁡x−1

Question

Evaluate the integral.  ∫dxcos⁡x−1

porschomcl · Accepted Answer

Step 1  to evaluate: ∫dxcos⁡x−1  Solution:  We have formula:  1−cos⁡2x=2sin2x  Therefore,  1−cos⁡x=2sin2(x2)  And also we have ∫csc2xdx=−cot⁡x  Step 2  Evaluate the above integral:  ∫dxcos⁡x−1=∫1−(1−cos⁡x)dx  =∫1−sin2(x2)dx  =∫−csc2(x2)dx  =cot⁡(x2)(12)+c   (Using ∫−csc2x=cot⁡x)  =2cot⁡(x2)+c  Hence, required answer is ∫dxcos⁡x−1=2cot⁡(x2)+c.

Nadine Salcido · Accepted Answer

∫1cos⁡x−1dx  Multiply both the numerator and the denominator by cos⁡x+1  =∫1cos⁡x−1⋅cos⁡x+1cos⁡x+1dx  =∫cos⁡x+1cos2x−1dx  =∫cos⁡x+1−sin2xdx  =∫−cos⁡xsin2x−1sin2xdx  =∫−cos⁡xsin⁡x⋅1sin⁡x−1sin2xdx  =∫−csc⁡xcot⁡x−csc2xdx  The antiderivative of csc⁡xcot⁡x and csc2x are −csc⁡x and −cot⁡x respectively.  =csc⁡x+cot⁡x+C  Result:  csc⁡x+cot⁡x+C

karton · Accepted Answer

Step 1I=∫dxcos⁡x−1Multiply and divide by (cos⁡x+1)I=∫dxcos⁡x−1∗cos⁡x+1cos⁡x+1=∫cos⁡x+1cos2⁡x−1dxWe know that sin2⁡x+cos2⁡x=1⇒cos2⁡x−1=−sin2⁡xI=−∫cos⁡x+1sin2⁡xdx=−∫(cos⁡xsin2⁡x+1sin2⁡x)dx=−∫(cos⁡xsin⁡x∗sin⁡x+1sin2⁡x)dx=−∫(cot⁡xcsc⁡x+csc2⁡x)dx=∫−cot⁡xcsc⁡xdx−∫csc2⁡xdx=csc⁡x+cot⁡x+cResult:∫dxcos⁡x−1=csc⁡x+cot⁡x+c

Evaluate the integral. \int \frac{dx}{\cos x-1}

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