Exams Joint Entrance Examination Mathematics Indefinite Integrals Question 338 Question 338 of 416 Medium MCQ
The value of 2 ∫ sin x d x sin ( x − π 4 ) \sqrt{2}\displaystyle\int \dfrac{\sin x\,dx}{\sin\left(x-\dfrac{\pi}{4}\right)} 2 ∫ sin ( x − 4 π ) sin x d x is:
Solutions · 1 Post your solution Official solution
Let x − π 4 = t ⇒ d x = d t x-\dfrac{\pi}{4}=t \Rightarrow dx=dt x − 4 π = t ⇒ d x = d t
I = 2 ∫ sin ( t + π 4 ) sin t d t = 2 ∫ 1 2 ( sin t + cos t ) sin t d t = ∫ ( 1 + cot t ) d t I = \sqrt{2}\int \dfrac{\sin\left(t+\dfrac{\pi}{4}\right)}{\sin t}\,dt = \sqrt{2}\int \dfrac{\dfrac{1}{\sqrt{2}}(\sin t+\cos t)}{\sin t}\,dt = \int(1+\cot t)\,dt I = 2 ∫ sin t sin ( t + 4 π ) d t = 2 ∫ sin t 2 1 ( sin t + cos t ) d t = ∫ ( 1 + cot t ) d t
= t + log ∣ sin t ∣ + c 1 = ( x − π 4 ) + log ∣ sin ( x − π 4 ) ∣ + c 1 = t+\log|\sin t|+c_1 = \left(x-\dfrac{\pi}{4}\right)+\log\left|\sin\left(x-\dfrac{\pi}{4}\right)\right|+c_1 = t + log ∣ sin t ∣ + c 1 = ( x − 4 π ) + log sin ( x − 4 π ) + c 1
= x + log ∣ sin ( x − π 4 ) ∣ + c = x+\log\left|\sin\left(x-\dfrac{\pi}{4}\right)\right|+c = x + log sin ( x − 4 π ) + c , where c = c 1 − π 4 c=c_1-\dfrac{\pi}{4} c = c 1 − 4 π Answer the question to unlock