Class 12 Maths Chapter 7 Integrals – Exercise 7.4 NCERT Solutions

Introduction

Exercise 7.4 focuses on integration by parts. Students learn how to integrate products of two functions using the formula and apply it to algebraic, trigonometric, exponential, and logarithmic functions. This technique is essential for solving complex integrals in calculus.

Formula Used

Integration by Parts:

$$\int udv=uv-\int vdu$$

where $u$ and $v$ are differentiable functions.

ILATE Rule (for choosing $u$):

• I → Inverse trigonometric

• L → Logarithmic

• A → Algebraic

• T → Trigonometric

• E → Exponential

Choose $u$ according to priority above.

Students Frequently Make Mistakes

• Choosing wrong function as $u$.

• Forgetting to apply ILATE rule.

• Errors in differentiating $u$ or integrating $dv$.

• Missing negative signs in trigonometric integrals.

• Forgetting constant of integration $+C$.

NCERT Questions with Step‑by‑Step Solutions (10)

Q1. Evaluate $\int x{e}^{x}dx$. Let $u=x,dv=e^{x}dx$.

$$du=dx,v=e^x$$

$$\int x{e}^{x}dx=x{e}^x-\int e^{x}dx=x{e}^x-e^x+C$$

Q2. Evaluate $\int x\sin xdx$. Let $u=x,dv=\sin xdx$.

$$du=dx,v=-\cos x$$

$$\int x\sin xdx=-x\cos x+\int \cos xdx=-x\cos x+\sin x+C$$

Q3. Evaluate $\int x\cos xdx$. Let $u=x,dv=\cos xdx$.

$$du=dx,v=\sin x$$

$$\int x\cos xdx=x\sin x-\int \sin xdx=x\sin x+\cos x+C$$

Q4. Evaluate $\int x\ln xdx$. Let $u=\ln x,dv=xdx$.

$$du=\frac{1}{x}dx,v=\frac{x^2}{2}$$

$$\int x\ln xdx=\frac{x^2}{2}\ln x-\int \frac{x^2}{2}\cdot \frac{1}{x}dx$$

$$=\frac{x^2}{2}\ln x-\frac{1}{2}\int xdx=\frac{x^2}{2}\ln x-\frac{x^2}{4}+C$$

Q5. Evaluate $\int e^x\cos xdx$. Let $u=\cos x,dv=e^{x}dx$.

$$du=-\sin xdx,v=e^x$$

$$\int e^x\cos xdx=e^x\cos x+\int e^x\sin xdx$$

Now apply by parts again → final result:

$$\frac{e^x}{2}(\sin x+\cos x)+C$$

Q6. Evaluate $\int e^x\sin xdx$. Similar process →

$$\frac{e^x}{2}(\sin x-\cos x)+C$$

Q7. Evaluate $\int x{e}^{2x}dx$. Let $u=x,dv=e^{2x}dx$.

$$du=dx,v=\frac{e^{2x}}{2}$$

$$\int x{e}^{2x}dx=\frac{x{e}^{2x}}{2}-\frac{1}{2}\int e^{2x}dx=\frac{x{e}^{2x}}{2}-\frac{e^{2x}}{4}+C$$

Q8. Evaluate $\int x{\tan }^{-1}xdx$. Let $u={\tan }^{-1}x,dv=xdx$.

$$du=\frac{1}{1+x^2}dx,v=\frac{x^2}{2}$$

$$\int x{\tan }^{-1}xdx=\frac{x^2}{2}{\tan }^{-1}x-\frac{1}{2}\int \frac{x^2}{1+x^2}dx$$

Simplify:

$$=\frac{x^2}{2}{\tan }^{-1}x-\frac{1}{2}\int (1-\frac{1}{1+x^2})dx$$

$$=\frac{x^2}{2}{\tan }^{-1}x-\frac{x}{2}+\frac{1}{2}{\tan }^{-1}x+C$$

Q9. Evaluate $\int x^2{e}^{x}dx$. Let $u=x^2,dv=e^{x}dx$.

$$du=2xdx,v=e^x$$

$$\int x^2{e}^{x}dx=x^2{e}^x-\int 2x{e}^{x}dx$$

Now apply by parts again →

$$=x^2{e}^x-(2x{e}^x-2e^x)+C=(x^2-2x+2)e^x+C$$

Q10. Evaluate $\int \ln xdx$. Let $u=\ln x,dv=dx$.

$$du=\frac{1}{x}dx,v=x$$

$$\int \ln xdx=x\ln x-\int 1dx=x\ln x-x+C$$

FAQs (10)

FAQ1. What is integration by parts? Method to integrate product of two functions.

FAQ2. What is ILATE rule? Priority rule for choosing $u$.

FAQ3. Why choose $u$ carefully? Wrong choice complicates integral.

FAQ4. What is integral of $x{e}^x$? $x{e}^x-e^x+C$.

FAQ5. What is integral of $x\sin x$? $-x\cos x+\sin x+C$.

FAQ6. What is integral of $x\cos x$? $x\sin x+\cos x+C$.

FAQ7. What is integral of $x\ln x$? $\frac{x^2}{2}\ln x-\frac{x^2}{4}+C$.

FAQ8. What is integral of $e^x\cos x$? $\frac{e^x}{2}(\sin x+\cos x)+C$.

FAQ9. What is integral of $e^x\sin x$? $\frac{e^x}{2}(\sin x-\cos x)+C$.

FAQ10. Why is Exercise 7.4 important? It builds foundation for advanced integration techniques.

Conclusion

Exercise 7.4 has 10 solved questions and 10 FAQs that strengthen your understanding of integration by parts. This builds the foundation for advanced calculus in Class 12 Maths.

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