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Q27:
Evaluate the definite integrals in Exercises 24 to 31.
27. $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x + \cos x}{\sqrt{\sin 2x}} dx$
Solution :
Step 1: Analyzing the Integrand and Substitution
Given the integral:
$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x + \cos x}{\sqrt{\sin 2x}} \, dx$
To simplify the denominator $\sqrt{\sin 2x}$, we express the numerator as the derivative of a function related to the denominator. We note that:
$(\sin x - \cos x)^2 = \sin^2 x + \cos^2 x - 2\sin x \cos x = 1 - \sin 2x$
Rearranging this, we get $\sin 2x = 1 - (\sin x - \cos x)^2$. [By the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ and the double angle formula $\sin 2x = 2\sin x \cos x$].
Step 2: Performing the Variable Substitution
Let $u = \sin x - \cos x$.
Then, the differential $du = (\cos x - (-\sin x)) \, dx = (\cos x + \sin x) \, dx$.
This perfectly matches the numerator of the integrand.
Step 3: Calculating New Limits of Integration
We transform the bounds of $x$ to bounds of $u$ using $u = \sin x - \cos x$:
- When $x = \frac{\pi}{6}$: $u = \sin\left(\frac{\pi}{6}\right) - \cos\left(\frac{\pi}{6}\right) = \frac{1}{2} - \frac{\sqrt{3}}{2} = \frac{1 - \sqrt{3}}{2}$
- When $x = \frac{\pi}{3}$: $u = \sin\left(\frac{\pi}{3}\right) - \cos\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} - \frac{1}{2} = \frac{\sqrt{3} - 1}{2}$
Step 4: Setting up the Transformed Integral
$I = \int_{\frac{1-\sqrt{3}}{2}}^{\frac{\sqrt{3}-1}{2}} \frac{du}{\sqrt{1 - u^2}}$
The standard integral formula is $\int \frac{du}{\sqrt{1 - u^2}} = \arcsin(u) + C$. [Per the fundamental antiderivative formula for inverse trigonometric functions].
Step 5: Evaluating the Definite Integral
$I = \left[ \arcsin(u) \right]_{\frac{1-\sqrt{3}}{2}}^{\frac{\sqrt{3}-1}{2}}$
$I = \arcsin\left(\frac{\sqrt{3}-1}{2}\right) - \arcsin\left(\frac{1-\sqrt{3}}{2}\right)$
Since $\arcsin(-x) = -\arcsin(x)$ for all $x \in [-1, 1]$, we have:
$I = \arcsin\left(\frac{\sqrt{3}-1}{2}\right) - \left(-\arcsin\left(\frac{\sqrt{3}-1}{2}\right)\right)$
$I = 2 \arcsin\left(\frac{\sqrt{3}-1}{2}\right)$
Final Solution: 2 \arcsin\left(\frac{\sqrt{3}-1}{2}\right)
More Questions from Class 12 Mathematics Integrals Miscellaneous Exercise on Chapter 7
- Q1: Integrate the functions in Exercises 1 to 23. 1. $\frac{1}{x-x^3}$
- Q2: Integrate the functions in Exercises 1 to 23. 2. $\frac{1}{\sqrt{x+a} + \sqrt{x+b}}$
- Q3: Integrate the functions in Exercises 1 to 23. 3. $\frac{1}{x\sqrt{ax-x^2}}$ [Hint: Put $x = \frac{a}{t}$]
- Q4: Integrate the functions in Exercises 1 to 23. 4. $\frac{1}{x^2(x^4+1)^{\frac{3}{4}}}$
- Q5: Integrate the functions in Exercises 1 to 23. 5. $\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}$ [Hint: $\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}} = \frac{1}{x^{\frac{1}{3}}(1+x^{\frac{1}{6}})}$, put $x=t^6$]
- Q6: Integrate the functions in Exercises 1 to 23. 6. $\frac{5x}{(x+1)(x^2+9)}$
- Q7: Integrate the functions in Exercises 1 to 23. 7. $\frac{\sin x}{\sin(x-a)}$
- Q8: Integrate the functions in Exercises 1 to 23. 8. $\frac{e^{5\log x} - e^{4\log x}}{e^{3\log x} - e^{2\log x}}$
- Q9: Integrate the functions in Exercises 1 to 23. 9. $\frac{\cos x}{\sqrt{4-\sin^2 x}}$
- Q10: Integrate the functions in Exercises 1 to 23. 10. $\frac{\sin^8 x - \cos^8 x}{1 - 2\sin^2 x \cos^2 x}$
- Q11: Integrate the functions in Exercises 1 to 23. 11. $\frac{1}{\cos(x+a)\cos(x+b)}$
- Q12: Integrate the functions in Exercises 1 to 23. 12. $\frac{x^3}{\sqrt{1-x^8}}$
- Q13: Integrate the functions in Exercises 1 to 23. 13. $\frac{e^x}{(1+e^x)(2+e^x)}$
- Q14: Integrate the functions in Exercises 1 to 23. 14. $\frac{1}{(x^2+1)(x^2+4)}$
- Q15: Integrate the functions in Exercises 1 to 23. 15. $\cos^3 x e^{\log \sin x}$
- Q16: Integrate the functions in Exercises 1 to 23. 16. $e^{3\log x} (x^4+1)^{-1}$
- Q17: Integrate the functions in Exercises 1 to 23. 17. $f'(ax+b)[f(ax+b)]^n$
- Q18: Integrate the functions in Exercises 1 to 23. 18. $\frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}}$
- Q19: Integrate the functions in Exercises 1 to 23. 19. $\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$
- Q20: Integrate the functions in Exercises 1 to 23. 20. $\frac{2+\sin 2x}{1+\cos 2x} e^x$
- Q21: Integrate the functions in Exercises 1 to 23. 21. $\frac{x^2+x+1}{(x+1)^2(x+2)}$
- Q22: Integrate the functions in Exercises 1 to 23. 22. $\tan^{-1} \sqrt{\frac{1-x}{1+x}}$
- Q23: Integrate the functions in Exercises 1 to 23. 23. $\frac{\sqrt{x^2+1}[\log(x^2+1)-2\log x]}{x^4}$
- Q24: Evaluate the definite integrals in Exercises 24 to 31. 24. $\int_{\frac{\pi}{2}}^{\pi} e^x (\frac{1-\sin x}{1-\cos x}) dx$
- Q25: Evaluate the definite integrals in Exercises 24 to 31. 25. $\int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos^4 x + \sin^4 x} dx$
- Q26: Evaluate the definite integrals in Exercises 24 to 31. 26. $\int_{0}^{\frac{\pi}{2}} \frac{\cos^2 x dx}{\cos^2 x + 4\sin^2 x}$
- Q28: Evaluate the definite integrals in Exercises 24 to 31. 28. $\int_{0}^{1} \frac{dx}{\sqrt{1+x}-\sqrt{x}}$
- Q29: Evaluate the definite integrals in Exercises 24 to 31. 29. $\int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{9+16\sin 2x} dx$
- Q30: Evaluate the definite integrals in Exercises 24 to 31. 30. $\int_{0}^{\frac{\pi}{2}} \sin 2x \tan^{-1}(\sin x) dx$
- Q31: Evaluate the definite integrals in Exercises 24 to 31. 31. $\int_{1}^{4} [|x-1|+|x-2|+|x-3|] dx$
- Q32: Prove the following (Exercises 32 to 37) 32. $\int_{1}^{3} \frac{dx}{x^2(x+1)} = \frac{2}{3} + \log \frac{2}{3}$
- Q33: Prove the following (Exercises 32 to 37) 33. $\int_{0}^{1} x e^x dx = 1$
- Q34: Prove the following (Exercises 32 to 37) 34. $\int_{-1}^{1} x^{17} \cos^4 x dx = 0$
- Q35: Prove the following (Exercises 32 to 37) 35. $\int_{0}^{\frac{\pi}{2}} \sin^3 x dx = \frac{2}{3}$
- Q36: Prove the following (Exercises 32 to 37) 36. $\int_{0}^{\frac{\pi}{4}} 2\tan^3 x dx = 1 - \log 2$
- Q37: Prove the following (Exercises 32 to 37) 37. $\int_{0}^{1} \sin^{-1} x dx = \frac{\pi}{2} - 1$
- Q38: Choose the correct answers in Exercises 38 to 40 38. $\int \frac{dx}{e^x + e^{-x}}$ is equal to
- Q39: Choose the correct answers in Exercises 38 to 40 39. $\int \frac{\cos 2x}{(\sin x + \cos x)^2} dx$ is equal to
- Q40: Choose the correct answers in Exercises 38 to 40 40. If $f(a+b-x) = f(x)$, then $\int_{a}^{b} x f(x) dx$ is equal to
CBSE Solutions for Class 12 Mathematics Integrals
- Integrals Class 12 EXERCISE 7.1
- Integrals Class 12 EXERCISE 7.10
- Integrals Class 12 EXERCISE 7.2
- Integrals Class 12 EXERCISE 7.3
- Integrals Class 12 EXERCISE 7.4
- Integrals Class 12 EXERCISE 7.5
- Integrals Class 12 EXERCISE 7.6
- Integrals Class 12 EXERCISE 7.7
- Integrals Class 12 EXERCISE 7.8
- Integrals Class 12 EXERCISE 7.9
Chapters in CBSE - Class 12 Mathematics
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