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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 20 Maths Textbook Solution.

Answers (1)

Answer: -\cos x \log (\cos x)+\cos x+c

Hint: Consider the first function as log(cos x) and second function as sin x

Given: Let I=\int \sin x \log (\cos x) d x

Solution: I=\int \sin x \log (\cos x) d x

                  \begin{aligned} &=\log (\cos x) \int \sin x d x-\int\left(\frac{d}{d x} \log (\cos x) \int \sin x d x\right) d x \\ &=-\cos x \log (\cos x)+\int \frac{-\sin x}{\cos x} \cos x d x \end{aligned}

So we get,

             =-\cos x \log (\cos x)-\int \sin x d x

Again by integrating the second term

           =-\cos x \log (\cos x)+\cos x+c


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