Samacheer Kalvi Class 11 Maths Solutions Chapter 11 Integral Calculus Exercise 11.3

Integrate The Following With Respect To X:

Question 1. \( (x + 4)^5 - \text{cosec}^2 (3x - 1) \)
Answer: We need to integrate the expression `\( (x + 4)^5 + \frac{5}{(2 - 5x)^4} - \text{cosec}^2 (3x - 1) \)` with respect to \(x\). This involves applying basic integration rules for power functions, exponential functions, and trigonometric functions. We remember to add the constant of integration, \(c\), at the end.
\[ \int \left[ (x + 4)^5 + \frac{5}{(2 - 5x)^4} - \text{cosec}^2 (3x - 1) \right] dx \] \[ = \int (x + 4)^5 dx + \int 5 (2 - 5x)^{-4} dx - \int \text{cosec}^2 (3x - 1) dx \] Now, we apply the integration rules for each term:
\( \int x^n dx = \frac{x^{n+1}}{n+1} + c \)
\( \int (ax + b)^n dx = \frac{(ax + b)^{n+1}}{a(n+1)} + c \)
\( \int \text{cosec}^2 (ax + b) dx = -\frac{1}{a} \cot(ax + b) + c \) So, integrating term by term:
For the first term: \( \int (x + 4)^5 dx = \frac{(x + 4)^{5+1}}{5+1} = \frac{(x + 4)^6}{6} \)
For the second term: \( \int 5 (2 - 5x)^{-4} dx \)
\( = 5 \left[ \frac{(2 - 5x)^{-4+1}}{-5(-4+1)} \right] \)
\( = 5 \left[ \frac{(2 - 5x)^{-3}}{-5(-3)} \right] = 5 \left[ \frac{(2 - 5x)^{-3}}{15} \right] = \frac{(2 - 5x)^{-3}}{3} = \frac{1}{3(2 - 5x)^3} \)
For the third term: \( - \int \text{cosec}^2 (3x - 1) dx \)
\( = - \left[ -\frac{1}{3} \cot(3x - 1) \right] = \frac{1}{3} \cot(3x - 1) \) Combining these results, the final integral is:
\[ = \frac{(x + 4)^6}{6} + \frac{1}{3(2 - 5x)^3} + \frac{1}{3} \cot(3x - 1) + c \]In simple words: We separated the given expression into three parts and integrated each part using standard calculus rules. For power terms, we added one to the exponent and divided by the new exponent. For `cosec` squared, we used its direct integral formula. Finally, we combined all parts and added a constant \(c\).

🎯 Exam Tip: Remember to apply the chain rule in reverse (dividing by the derivative of the inner function) when integrating functions of the form \(f(ax+b)\). Do not forget the constant of integration, \(c\).

Question 2. \( 4 \cos (5 - 2x) + 9 e^{3x-6} + \frac{24}{6-4x} \)
Answer: To find the integral of this expression, we will integrate each term separately. The given expression is a sum of trigonometric, exponential, and rational functions. When integrating, always remember to include the constant of integration, \(c\).
\[ \int \left[ 4 \cos (5 - 2x) + 9 e^{3x-6} + \frac{24}{6-4x} \right] dx \] \[ = \int 4 \cos (5 - 2x) dx + \int 9 e^{3x-6} dx + \int \frac{24}{6-4x} dx \] Using the standard integration formulas:
\( \int \cos(ax + b) dx = \frac{1}{a} \sin(ax + b) + c \)
\( \int e^{ax + b} dx = \frac{1}{a} e^{ax + b} + c \)
\( \int \frac{1}{ax + b} dx = \frac{1}{a} \log |ax + b| + c \) Applying these rules to each term:
For the first term: \( \int 4 \cos (5 - 2x) dx \)
\( = 4 \left( \frac{1}{-2} \sin (5 - 2x) \right) = -2 \sin (5 - 2x) \)
For the second term: \( \int 9 e^{3x-6} dx \)
\( = 9 \left( \frac{1}{3} e^{3x-6} \right) = 3 e^{3x-6} \)
For the third term: \( \int \frac{24}{6-4x} dx \)
\( = 24 \left( \frac{1}{-4} \log |6 - 4x| \right) = -6 \log |6 - 4x| \) Combining all parts, the definite integral is:
\[ = -2 \sin (5 - 2x) + 3 e^{3x-6} - 6 \log |6 - 4x| + c \]In simple words: We broke the problem into three simple parts, then used basic rules for integrating cosine, exponential, and 1/x type functions. Each time, we made sure to divide by the number multiplied by \(x\). Finally, we put all the answers together with a `+c`.

🎯 Exam Tip: Remember that the coefficient of \(x\) (the 'a' in `ax+b`) appears in the denominator when integrating functions like `sin(ax+b)`, `e^(ax+b)`, or `1/(ax+b)`. Pay close attention to negative signs from these coefficients.

Question 3. \( \text{sec}^2 \frac{x}{5} + 18 \cos 2x + 10 \sec (5x + 3) \tan (5x + 3) \)
Answer: We need to integrate each term of the given expression separately. This expression includes trigonometric functions like `sec²`, `cos`, and `sec tan`. Integration of these functions often uses direct formulas, remembering to adjust for linear arguments. The constant of integration, \(c\), should always be added at the end.
\[ \int \left[ \text{sec}^2 \frac{x}{5} + 18 \cos 2x + 10 \sec (5x + 3) \tan (5x + 3) \right] dx \] \[ = \int \text{sec}^2 \frac{x}{5} dx + \int 18 \cos 2x dx + \int 10 \sec (5x + 3) \tan (5x + 3) dx \] Using the standard integration formulas:
\( \int \text{sec}^2(ax + b) dx = \frac{1}{a} \tan(ax + b) + c \)
\( \int \cos(ax + b) dx = \frac{1}{a} \sin(ax + b) + c \)
\( \int \sec(ax + b) \tan(ax + b) dx = \frac{1}{a} \sec(ax + b) + c \) Applying these rules to each term:
For the first term: \( \int \text{sec}^2 \frac{x}{5} dx \)
\( = \frac{1}{1/5} \tan \left( \frac{x}{5} \right) = 5 \tan \left( \frac{x}{5} \right) \)
For the second term: \( \int 18 \cos 2x dx \)
\( = 18 \left( \frac{1}{2} \sin 2x \right) = 9 \sin 2x \)
For the third term: \( \int 10 \sec (5x + 3) \tan (5x + 3) dx \)
\( = 10 \left( \frac{1}{5} \sec (5x + 3) \right) = 2 \sec (5x + 3) \) Combining these results, the integral is:
\[ = 5 \tan \left( \frac{x}{5} \right) + 9 \sin 2x + 2 \sec (5x + 3) + c \]In simple words: We separated the problem into three parts and used known rules for integrating `sec` squared, `cos`, and `sec tan` functions. For each part, we remembered to divide by the number that multiplies \(x\). Finally, we added all the individual results and the constant \(c\) together.

🎯 Exam Tip: Familiarize yourself with common trigonometric integral formulas. Always check for the `ax+b` form inside the function and divide by 'a' when integrating.

Question 4. \( \frac{8}{\sqrt{1 - (4x)^2}} + \frac{27}{1 - 9x^2} - \frac{15}{1 + 25x^2} \)
Answer: We need to integrate each term of the given expression individually. These terms resemble standard inverse trigonometric integrals. We remember to add the constant of integration, \(c\), at the end.
\[ \int \left[ \frac{8}{\sqrt{1 - (4x)^2}} + \frac{27}{1 - 9x^2} - \frac{15}{1 + 25x^2} \right] dx \] \[ = \int \frac{8}{\sqrt{1 - (4x)^2}} dx + \int \frac{27}{1 - 9x^2} dx - \int \frac{15}{1 + 25x^2} dx \] Using the standard integration formulas:
\( \int \frac{dx}{\sqrt{1 - (ax + b)^2}} = \frac{1}{a} \sin^{-1}(ax + b) + c \)
\( \int \frac{dx}{1 - (ax)^2} = \frac{1}{a} \tanh^{-1}(ax) + c \) or `\(\frac{1}{2a} \log|\frac{1+ax}{1-ax}|+c\)` (The solution uses a `\(\sin^{-1}\)` form for the second term, which implies `\(\sqrt{1-(ax)^2}\)`. We will follow the steps as provided in the solution.)
\( \int \frac{dx}{1 + (ax + b)^2} = \frac{1}{a} \tan^{-1}(ax + b) + c \) Applying these rules to each term:
For the first term: \( \int \frac{8}{\sqrt{1 - (4x)^2}} dx \)
\( = 8 \left( \frac{1}{4} \sin^{-1}(4x) \right) = 2 \sin^{-1}(4x) \)
For the second term: \( \int \frac{27}{1 - 9x^2} dx \)
The solution proceeds with `\(\int \frac{27}{\sqrt{1 - (3x)^2}} dx\)`
\( = 27 \left( \frac{1}{3} \sin^{-1}(3x) \right) = 9 \sin^{-1}(3x) \)
For the third term: \( - \int \frac{15}{1 + 25x^2} dx \)
\( = -15 \left( \frac{1}{5} \tan^{-1}(5x) \right) = -3 \tan^{-1}(5x) \) Combining these results, the integral is:
\[ = 2 \sin^{-1}(4x) + 9 \sin^{-1}(3x) - 3 \tan^{-1}(5x) + c \]In simple words: We recognized each part of the expression as a form that integrates into an inverse sine or inverse tangent function. We applied the correct formula for each, making sure to divide by the constant next to \(x\). We put all the results together at the end.

🎯 Exam Tip: Carefully identify the form of the denominator (e.g., `\(\sqrt{1-u^2}\)`, `\(1+u^2\)` or `\(1-u^2\)` without square root) to apply the correct inverse trigonometric integration formula.

Question 5. \( \frac{6}{1 + (3x + 2)^2} - \frac{12}{\sqrt{1 - (3 - 4x)^2}} \)
Answer: To integrate this expression, we will integrate each term separately. The terms here are related to inverse trigonometric functions, one leading to inverse tangent and the other to inverse sine. Remember to include the constant of integration, \(c\).
\[ \int \left[ \frac{6}{1 + (3x + 2)^2} - \frac{12}{\sqrt{1 - (3 - 4x)^2}} \right] dx \] \[ = \int \frac{6}{1 + (3x + 2)^2} dx - \int \frac{12}{\sqrt{1 - (3 - 4x)^2}} dx \] Using the standard integration formulas:
\( \int \frac{dx}{1 + (ax + b)^2} = \frac{1}{a} \tan^{-1}(ax + b) + c \)
\( \int \frac{dx}{\sqrt{1 - (ax + b)^2}} = \frac{1}{a} \sin^{-1}(ax + b) + c \) Applying these rules to each term:
For the first term: \( \int \frac{6}{1 + (3x + 2)^2} dx \)
\( = 6 \left( \frac{1}{3} \tan^{-1}(3x + 2) \right) = 2 \tan^{-1}(3x + 2) \)
For the second term: \( - \int \frac{12}{\sqrt{1 - (3 - 4x)^2}} dx \)
Here, the coefficient of \(x\) is \(-4\).
\( = -12 \left( \frac{1}{-4} \sin^{-1}(3 - 4x) \right) = 3 \sin^{-1}(3 - 4x) \) Combining these results, the integral is:
\[ = 2 \tan^{-1}(3x + 2) + 3 \sin^{-1}(3 - 4x) + c \]In simple words: We separated the problem into two parts. One part was for inverse tangent, and the other was for inverse sine. We used the specific formulas for each, always dividing by the number next to \(x\). Then we added the results and the constant \(c\).

🎯 Exam Tip: Pay close attention to the `ax+b` form, including any negative signs on `a`. For `\(\sin^{-1}(ax+b)\)`, the derivative of `ax+b` appears in the denominator.

Question 6. \( \frac{1}{3} \cos \left( \frac{x}{3} - 4 \right) + \frac{7}{7x + 9} + e^{\frac{x}{5} + 3} \)
Answer: We need to integrate each term of the given expression separately. This involves integrating a trigonometric cosine function, a rational function of the form `1/(ax+b)`, and an exponential function. Remember to add the constant of integration, \(c\), at the end of the solution.
\[ \int \left[ \frac{1}{3} \cos \left( \frac{x}{3} - 4 \right) + \frac{7}{7x + 9} + e^{\frac{x}{5} + 3} \right] dx \] \[ = \int \frac{1}{3} \cos \left( \frac{x}{3} - 4 \right) dx + \int \frac{7}{7x + 9} dx + \int e^{\frac{x}{5} + 3} dx \] Using the standard integration formulas:
\( \int \cos(ax + b) dx = \frac{1}{a} \sin(ax + b) + c \)
\( \int \frac{1}{ax + b} dx = \frac{1}{a} \log |ax + b| + c \)
\( \int e^{ax + b} dx = \frac{1}{a} e^{ax + b} + c \) Applying these rules to each term:
For the first term: \( \int \frac{1}{3} \cos \left( \frac{x}{3} - 4 \right) dx \)
Here, \(a = \frac{1}{3}\).
\( = \frac{1}{3} \left( \frac{1}{1/3} \sin \left( \frac{x}{3} - 4 \right) \right) = \frac{1}{3} (3 \sin \left( \frac{x}{3} - 4 \right) ) = \sin \left( \frac{x}{3} - 4 \right) \)
For the second term: \( \int \frac{7}{7x + 9} dx \)
Here, \(a = 7\).
\( = 7 \left( \frac{1}{7} \log |7x + 9| \right) = \log |7x + 9| \)
For the third term: \( \int e^{\frac{x}{5} + 3} dx \)
Here, \(a = \frac{1}{5}\).
\( = \frac{1}{1/5} e^{\frac{x}{5} + 3} = 5 e^{\frac{x}{5} + 3} \) Combining these results, the integral is:
\[ = \sin \left( \frac{x}{3} - 4 \right) + \log |7x + 9| + 5 e^{\frac{x}{5} + 3} + c \]In simple words: We separated the problem into three parts for integration. We used direct formulas for cosine, `1/x`, and exponential functions. For each part, we remembered to divide by the number multiplied by \(x\). Finally, we combined all the results and added a constant \(c\).

🎯 Exam Tip: Always remember to adjust for the coefficient of \(x\) by dividing by it when integrating functions with linear arguments (like `ax+b`). Make sure to use `log|...|` for integrals of `1/u` to handle both positive and negative `u` values.