Integrate ∫dxsin3xsin⁡(x+α)∫1sin3xsin⁡(x+α)dx=∫1sin⁡xsin⁡xsin⁡(x+α)dx=1sin⁡α∫sin⁡(x+α−x)sin⁡xsin⁡xsin⁡(x+α)dx=1sin⁡α∫sin⁡(x+α)cos⁡x−cos⁡(x+α)sin⁡{x}sin⁡xsin⁡xsin⁡(x+α)dx=1sin⁡α∫sin⁡(x+α)cos⁡xsin⁡xsin⁡xsin⁡(x+α)dx−1sin⁡α∫cos⁡(x+α)sin⁡{x}sin⁡xsin⁡xsin⁡(x+α)dxIs it possible to proceed further and complete the integration or what is the right substitution to find the solution ?

Question

hadnya1qd · Accepted Answer

Divide by sin2x on numerator and denominator and use the identitysin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B  to get to following:∫csc2xdxcos⁡a+sin⁡acot⁡xNoting that cot⁡′(x)=−csc2x it is straightforward now. The final antiderivative is:−2cos⁡a+sin⁡acot⁡xsin⁡a+c

Integrate \int \frac{dx}{\sqrt{\sin^3x\sin(x+\alpha)}} \int \frac{1}{\sqrt{\sin^3x\sin(x+\alpha)}}dx=\int \frac{1}{\sin x\sqrt{\sin x\sin(x+\alpha)}}dx =\frac{1}{\sin\alpha}\int \frac{\sin(x+\alpha-x)}{\sin x\sqrt{\sin x\sin(x+\alpha)}}dx=\frac{1}{\sin\alpha}\int\frac{\sin(x+\alpha)\cos

Answered question

Answer & Explanation

New Questions in Integral Calculus