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f(x)=(cos(x)+sin(x))/sin(2x)−6−5−4−3−2−10123456−3-2.5−2-1.5−1-0.500.511.522.53xy

Exercise

∫cos⁡(x)+sin⁡(x)sin⁡(2x)dx\int\frac{\cos\left(x\right)+\sin\left(x\right)}{\sin\left(2x\right)}dx

Step-by-step Solution

1

Expand the fraction cos⁡(x)+sin⁡(x)sin⁡(2x)\frac{\cos\left(x\right)+\sin\left(x\right)}{\sin\left(2x\right)} into 22 simpler fractions with common denominator sin⁡(2x)\sin\left(2x\right)

∫(cos⁡(x)sin⁡(2x)+sin⁡(x)sin⁡(2x))dx\int\left(\frac{\cos\left(x\right)}{\sin\left(2x\right)}+\frac{\sin\left(x\right)}{\sin\left(2x\right)}\right)dx
2

Simplify the expression

∫12sin⁡(x)dx+∫12cos⁡(x)dx\int\frac{1}{2\sin\left(x\right)}dx+\int\frac{1}{2\cos\left(x\right)}dx
3

The integral ∫12sin⁡(x)dx\int\frac{1}{2\sin\left(x\right)}dx results in: −12ln⁡(csc⁡(x)+cot⁡(x))-\frac{1}{2}\ln\left(\csc\left(x\right)+\cot\left(x\right)\right)

−12ln⁡(csc⁡(x)+cot⁡(x))-\frac{1}{2}\ln\left(\csc\left(x\right)+\cot\left(x\right)\right)
4

The integral ∫12cos⁡(x)dx\int\frac{1}{2\cos\left(x\right)}dx results in: 12ln⁡(sec⁡(x)+tan⁡(x))\frac{1}{2}\ln\left(\sec\left(x\right)+\tan\left(x\right)\right)

12ln⁡(sec⁡(x)+tan⁡(x))\frac{1}{2}\ln\left(\sec\left(x\right)+\tan\left(x\right)\right)
5

Gather the results of all integrals

−12ln⁡∣csc⁡(x)+cot⁡(x)∣+12ln⁡∣sec⁡(x)+tan⁡(x)∣-\frac{1}{2}\ln\left|\csc\left(x\right)+\cot\left(x\right)\right|+\frac{1}{2}\ln\left|\sec\left(x\right)+\tan\left(x\right)\right|
6

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration CC

−12ln⁡∣csc⁡(x)+cot⁡(x)∣+12ln⁡∣sec⁡(x)+tan⁡(x)∣+C0-\frac{1}{2}\ln\left|\csc\left(x\right)+\cot\left(x\right)\right|+\frac{1}{2}\ln\left|\sec\left(x\right)+\tan\left(x\right)\right|+C_0

Final answer to the exercise

−12ln⁡∣csc⁡(x)+cot⁡(x)∣+12ln⁡∣sec⁡(x)+tan⁡(x)∣+C0-\frac{1}{2}\ln\left|\csc\left(x\right)+\cot\left(x\right)\right|+\frac{1}{2}\ln\left|\sec\left(x\right)+\tan\left(x\right)\right|+C_0

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