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Q10:
Integrate the functions in Exercises 1 to 22.
10. $(\sin^{-1} x)^2$
Solution :
Step 1: Application of Integration by Parts
To evaluate the integral $I = \int (\sin^{-1} x)^2 \, dx$, we utilize the method of Integration by Parts: $\int u \, dv = uv - \int v \, du$. [Per the Integration by Parts Theorem].
Let $u = (\sin^{-1} x)^2$ and $dv = dx$.
Then, by the chain rule: $du = 2(\sin^{-1} x) \cdot \frac{1}{\sqrt{1-x^2}} \, dx$ and $v = x$.
Applying the formula:
$I = x(\sin^{-1} x)^2 - \int x \cdot \frac{2 \sin^{-1} x}{\sqrt{1-x^2}} \, dx$ $I = x(\sin^{-1} x)^2 - 2 \int \frac{x \sin^{-1} x}{\sqrt{1-x^2}} \, dx$Step 2: Substitution Method for the Secondary Integral
Let $J = \int \frac{x \sin^{-1} x}{\sqrt{1-x^2}} \, dx$. To solve $J$, we use the substitution method. Let $t = \sin^{-1} x$. Then $x = \sin t$ and $dx = \cos t \, dt$.
Substituting into the integral:
$J = \int \frac{\sin t \cdot t}{\sqrt{1-\sin^2 t}} \cdot \cos t \, dt$ $J = \int \frac{t \sin t}{\cos t} \cdot \cos t \, dt = \int t \sin t \, dt$Step 3: Evaluating the Integral of $t \sin t$
Using Integration by Parts again on $\int t \sin t \, dt$, let $u_1 = t$ and $dv_1 = \sin t \, dt$.
Then $du_1 = dt$ and $v_1 = -\cos t$.
$\int t \sin t \, dt = t(-\cos t) - \int (-\cos t) \, dt$ $\int t \sin t \, dt = -t \cos t + \sin t + C_1$Step 4: Back-substitution to the Original Variables
Since $t = \sin^{-1} x$, we have $\sin t = x$ and $\cos t = \sqrt{1-x^2}$.
$J = -(\sin^{-1} x)\sqrt{1-x^2} + x + C_1$Step 5: Final Assembly of the Solution
Substitute $J$ back into the original equation for $I$:
$I = x(\sin^{-1} x)^2 - 2 \left[ -(\sin^{-1} x)\sqrt{1-x^2} + x \right] + C$ $I = x(\sin^{-1} x)^2 + 2\sqrt{1-x^2} \sin^{-1} x - 2x + C$Final Solution: $x(\sin^{-1} x)^2 + 2\sqrt{1-x^2} \sin^{-1} x - 2x + C$
More Questions from Class 12 Mathematics Integrals EXERCISE 7.6
- Q1: Integrate the functions in Exercises 1 to 22. 1. $x \sin x$
- Q2: Integrate the functions in Exercises 1 to 22. 2. $x \sin 3x$
- Q3: Integrate the functions in Exercises 1 to 22. 3. $x^2 e^x$
- Q4: Integrate the functions in Exercises 1 to 22. 4. $x \log x$
- Q5: Integrate the functions in Exercises 1 to 22. 5. $x \log 2x$
- Q6: Integrate the functions in Exercises 1 to 22. 6. $x^2 \log x$
- Q7: Integrate the functions in Exercises 1 to 22. 7. $x \sin^{-1} x$
- Q8: Integrate the functions in Exercises 1 to 22. 8. $x \tan^{-1} x$
- Q9: Integrate the functions in Exercises 1 to 22. 9. $x \cos^{-1} x$
- Q11: Integrate the functions in Exercises 1 to 22. 11. $\frac{x \cos^{-1} x}{\sqrt{1-x^2}}$
- Q12: Integrate the functions in Exercises 1 to 22. 12. $x \sec^2 x$
- Q13: Integrate the functions in Exercises 1 to 22. 13. $\tan^{-1} x$
- Q14: Integrate the functions in Exercises 1 to 22. 14. $x (\log x)^2$
- Q15: Integrate the functions in Exercises 1 to 22. 15. $(x^2+1) \log x$
- Q16: Integrate the functions in Exercises 1 to 22. 16. $e^x(\sin x + \cos x)$
- Q17: Integrate the functions in Exercises 1 to 22. 17. $\frac{x e^x}{(1+x)^2}$
- Q18: Integrate the functions in Exercises 1 to 22. 18. $e^x (\frac{1+\sin x}{1+\cos x})$
- Q19: Integrate the functions in Exercises 1 to 22. 19. $e^x (\frac{1}{x} - \frac{1}{x^2})$
- Q20: Integrate the functions in Exercises 1 to 22. 20. $\frac{(x-3)e^x}{(x-1)^3}$
- Q21: Integrate the functions in Exercises 1 to 22. 21. $e^{2x} \sin x$
- Q22: Integrate the functions in Exercises 1 to 22. 22. $\sin^{-1} (\frac{2x}{1+x^2})$
- Q23: Choose the correct answer in Exercises 23 and 24. 23. $\int x^2 e^{x^3} dx$ equals
- Q24: Choose the correct answer in Exercises 23 and 24. 24. $\int e^x \sec x (1 + \tan x) dx$ equals
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 7.7
- Integrals Class 12 EXERCISE 7.8
- Integrals Class 12 EXERCISE 7.9
- Integrals Class 12 Miscellaneous Exercise on Chapter 7
Chapters in CBSE - Class 12 Mathematics
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