Exams Joint Entrance Examination Mathematics Indefinite Integrals Question 352 Question 352 of 416 Medium MCQ
∫ d x cos x − sin x \displaystyle\int \dfrac{dx}{\cos x-\sin x} ∫ cos x − sin x d x is equal to:
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∫ d x cos x − sin x = ∫ d x 2 ( 1 2 cos x − 1 2 sin x ) = 1 2 ∫ d x cos ( x + π 4 ) \int \dfrac{dx}{\cos x-\sin x} = \int \dfrac{dx}{\sqrt{2}\left(\dfrac{1}{\sqrt{2}}\cos x-\dfrac{1}{\sqrt{2}}\sin x\right)} = \dfrac{1}{\sqrt{2}}\int \dfrac{dx}{\cos\left(x+\dfrac{\pi}{4}\right)} ∫ cos x − sin x d x = ∫ 2 ( 2 1 cos x − 2 1 sin x ) d x = 2 1 ∫ cos ( x + 4 π ) d x
= 1 2 ∫ sec ( x + π 4 ) d x = \dfrac{1}{\sqrt{2}}\int \sec\left(x+\dfrac{\pi}{4}\right)dx = 2 1 ∫ sec ( x + 4 π ) d x
Since ∫ sec x d x = log ∣ tan ( π 4 + x 2 ) ∣ \int \sec x\,dx = \log\left|\tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)\right| ∫ sec x d x = log tan ( 4 π + 2 x ) :
= 1 2 log ∣ tan ( π 4 + x 2 + π 8 ) ∣ + C = 1 2 log ∣ tan ( x 2 + 3 π 8 ) ∣ + C = \dfrac{1}{\sqrt{2}}\log\left|\tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}+\dfrac{\pi}{8}\right)\right|+C = \dfrac{1}{\sqrt{2}}\log\left|\tan\left(\dfrac{x}{2}+\dfrac{3\pi}{8}\right)\right|+C = 2 1 log tan ( 4 π + 2 x + 8 π ) + C = 2 1 log tan ( 2 x + 8 3 π ) + C Answer the question to unlock