Question
Download Solution PDFWhat is \(\int\frac{(\cos x)^{1.5}−(\sin x)^{1.5}}{\sqrt{\sin x⋅\cos x}}\)dx equal to ?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven,
I = \(\int\frac{(\cos x)^{1.5}−(\sin x)^{1.5}}{\sqrt{\sin x⋅\cos x}}dx\)
⇒ I = \(\int\frac{\cos x}{\sqrt{\sin x}}- \frac{\sin x}{\sqrt{\cos x}}dx\)
⇒ I = \(\int\frac{\cos x}{\sqrt{\sin x}}dx- \int \frac{\sin x}{\sqrt{\cos x}}dx\) = I1 - I2 (let)
Computing I1,
I1 = \(\int\frac{\cos x}{\sqrt{\sin x}}dx\)
Put sin x = t → cos x dx = dt
⇒ I1 = \(\int\frac{dt}{\sqrt{t}} = {t^{-{1 \over 2}+1} \over{-{1 \over 2}+1}} + c_1\)
⇒ I1 = \(2\sqrt t + c_1\)
⇒ I1 = \(2\sqrt {\sin x} + c_1\)
Similarly,
I2 = \(\int\frac{\sin x}{\sqrt{\cos x}}dx\)
Put cos x = t → - sin x dx = dt
⇒ I2 = \(\int\frac{dt}{\sqrt{t}} = -{t^{-{1 \over 2}+1} \over{-{1 \over 2}+1}} + c_2\)
⇒ I2 = \(-2\sqrt t + c_2\)
⇒ I2 = \(-2\sqrt {\cos x} + c_2\)
Putting these value in I,
⇒ I = I1 - I2 = \(2\sqrt {\sin x} + c_1\) - \((-2\sqrt {\cos x}) + c_2\)
⇒ I = \(2\sqrt{\sin \text{x}}+2 \sqrt{\cos \text{x}}\) + c
∴ The correct answer is option (3).
Last updated on May 30, 2025
->UPSC has released UPSC NDA 2 Notification on 28th May 2025 announcing the NDA 2 vacancies.
-> A total of 406 vacancies have been announced for NDA 2 Exam 2025.
->The NDA exam date 2025 has been announced for cycle 2. The written examination will be held on 14th September 2025.
-> Earlier, the UPSC NDA 1 Exam Result has been released on the official website.
-> The selection process for the NDA exam includes a Written Exam and SSB Interview.
-> Candidates who get successful selection under UPSC NDA will get a salary range between Rs. 15,600 to Rs. 39,100.
-> Candidates must go through the NDA previous year question paper. Attempting the NDA mock test is also essential.