Samacheer Kalvi Class 11 Maths Solutions Chapter 11 Integral Calculus Exercise 11.9

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.9

Integrate the following with respect to x:

Question 1. \( e^x (\tan x + \log \sec x) \)
Answer: Let the integral be \( I \).
\( I = \int e^x (\tan x + \log \sec x) \, dx \)
We know the formula \( \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \).
Let \( f(x) = \log \sec x \).
Then, the derivative \( f'(x) = \frac{d}{dx} (\log \sec x) = \frac{1}{\sec x} \cdot (\sec x \tan x) = \tan x \).
We can see that \( f(x) = \log \sec x \) and \( f'(x) = \tan x \).
So, the integral fits the special form.
\( \implies I = e^x \log |\sec x| + C \)
This method makes solving certain complex integrals much simpler.
In simple words: This question asks us to integrate an expression. We use a special rule where if the expression has \( e^x \) multiplied by a function plus its derivative, the answer is just \( e^x \) times the original function.

🎯 Exam Tip: Always look for the \( e^x[f(x) + f'(x)] \) pattern when \( e^x \) is present in an integral, as it simplifies the solution significantly.

Question 2. \( e^x (\frac{x-1}{2x^2}) \)
Answer: Let the integral be \( I \).
\( I = \int e^x \left(\frac{x-1}{2x^2}\right) \, dx \)
We can split the fraction inside the integral:
\( \implies I = \frac{1}{2} \int e^x \left(\frac{x}{x^2} - \frac{1}{x^2}\right) \, dx \)
\( \implies I = \frac{1}{2} \int e^x \left(\frac{1}{x} - \frac{1}{x^2}\right) \, dx \)
This integral is in the form \( k \int e^x [f(x) + f'(x)] \, dx \).
Let \( f(x) = \frac{1}{x} \).
Then, the derivative \( f'(x) = \frac{d}{dx} \left(\frac{1}{x}\right) = -\frac{1}{x^2} \).
So, we have \( f(x) = \frac{1}{x} \) and \( f'(x) = -\frac{1}{x^2} \).
\( \implies I = \frac{1}{2} e^x \left(\frac{1}{x}\right) + C \)
Thus, the constant \( \frac{1}{2} \) is simply multiplied by the result of the special integration rule.
In simple words: We first break down the fraction into two parts. Then, we find a function whose derivative is the second part, similar to the previous problem. This lets us use the special \( e^x \) rule for integration.

🎯 Exam Tip: When faced with fractions inside an \( e^x \) integral, try to simplify or split the fraction to match the \( f(x) + f'(x) \) form. Sometimes factoring out a constant can also help.

Question 3. \( e^x \sec x (1 + \tan x) \)
Answer: Let the integral be \( I \).
\( I = \int e^x \sec x (1 + \tan x) \, dx \)
Distribute \( \sec x \) inside the parenthesis:
\( \implies I = \int e^x (\sec x + \sec x \tan x) \, dx \)
This integral is in the form \( \int e^x [f(x) + f'(x)] \, dx \).
Let \( f(x) = \sec x \).
Then, the derivative \( f'(x) = \frac{d}{dx} (\sec x) = \sec x \tan x \).
So, we have \( f(x) = \sec x \) and \( f'(x) = \sec x \tan x \).
\( \implies I = e^x \sec x + C \)
Recognizing standard derivatives like that of \( \sec x \) is key to quickly solving these problems.
In simple words: First, we multiply \( \sec x \) by both parts inside the bracket. This creates an expression where one part is the derivative of the other. Then, we use the special \( e^x \) rule to get the answer.

🎯 Exam Tip: Memorize common trigonometric derivatives, especially those for \( \sec x \), \( \tan x \), \( \cot x \), \( \csc x \) as they frequently appear in this type of integral.

Question 4. \( e^x \left(\frac{2 + \sin 2x}{1 + \cos 2x}\right) \)
Answer: Let the integral be \( I \).
\( I = \int e^x \left(\frac{2 + \sin 2x}{1 + \cos 2x}\right) \, dx \)
We use trigonometric identities: \( \sin 2x = 2 \sin x \cos x \) and \( 1 + \cos 2x = 2 \cos^2 x \).
\( \implies I = \int e^x \left(\frac{2 + 2 \sin x \cos x}{2 \cos^2 x}\right) \, dx \)
Factor out 2 from the numerator:
\( \implies I = \int e^x \left(\frac{2(1 + \sin x \cos x)}{2 \cos^2 x}\right) \, dx \)
Cancel out the 2s:
\( \implies I = \int e^x \left(\frac{1 + \sin x \cos x}{\cos^2 x}\right) \, dx \)
Split the fraction:
\( \implies I = \int e^x \left(\frac{1}{\cos^2 x} + \frac{\sin x \cos x}{\cos^2 x}\right) \, dx \)
Simplify using \( \frac{1}{\cos^2 x} = \sec^2 x \) and \( \frac{\sin x}{\cos x} = \tan x \):
\( \implies I = \int e^x (\sec^2 x + \tan x) \, dx \)
Rearrange the terms to fit the formula:
\( \implies I = \int e^x (\tan x + \sec^2 x) \, dx \)
This integral is in the form \( \int e^x [f(x) + f'(x)] \, dx \).
Let \( f(x) = \tan x \).
Then, the derivative \( f'(x) = \frac{d}{dx} (\tan x) = \sec^2 x \).
So, we have \( f(x) = \tan x \) and \( f'(x) = \sec^2 x \).
\( \implies I = e^x \tan x + C \)
This problem demonstrates how trigonometric identities are crucial for simplifying expressions before integration.
In simple words: We use math rules for sine and cosine to change the fraction into a simpler form. After that, we split the fraction into two parts: \( \sec^2 x \) and \( \tan x \). Since \( \sec^2 x \) is the derivative of \( \tan x \), we can use the special \( e^x \) rule to find the answer.

🎯 Exam Tip: Always be prepared to use fundamental trigonometric identities to simplify expressions, especially for integrals involving \( \sin 2x \) and \( \cos 2x \).

Question 5. \( e^{\tan^{-1} x} \left(\frac{1 + x + x^2}{1 + x^2}\right) \)
Answer: Let the integral be \( I \).
\( I = \int e^{\tan^{-1} x} \left(\frac{1 + x + x^2}{1 + x^2}\right) \, dx \)
Let's use substitution. Put \( t = \tan^{-1} x \).
This means \( x = \tan t \).
Differentiating both sides with respect to \( x \): \( \frac{dt}{dx} = \frac{1}{1+x^2} \), so \( dt = \frac{1}{1+x^2} \, dx \).
Now, rewrite the integral in terms of \( t \). We can split the original fraction:
\( \frac{1 + x + x^2}{1 + x^2} = \frac{(1+x^2) + x}{1+x^2} = 1 + \frac{x}{1+x^2} \).
However, the solution simplifies it as \( \frac{1 + x + x^2}{1 + x^2} = 1 + \frac{x}{1+x^2} \), which becomes part of the \( f(t) + f'(t) \) form after manipulation.
Let's follow the provided steps closely.
The substitution leads to:
\( I = \int e^t (1 + \tan t + \tan^2 t) \, dt \)
We know the identity \( 1 + \tan^2 t = \sec^2 t \).
\( \implies I = \int e^t (\sec^2 t + \tan t) \, dt \)
Rearrange the terms:
\( \implies I = \int e^t (\tan t + \sec^2 t) \, dt \)
This integral is in the form \( \int e^t [f(t) + f'(t)] \, dt \).
Let \( f(t) = \tan t \).
Then, the derivative \( f'(t) = \frac{d}{dt} (\tan t) = \sec^2 t \).
So, we have \( f(t) = \tan t \) and \( f'(t) = \sec^2 t \).
\( \implies I = e^t \tan t + C \)
Now, substitute back \( t = \tan^{-1} x \) and \( \tan t = x \):
\( \implies I = e^{\tan^{-1} x} \cdot x + C \)
\( \implies I = x e^{\tan^{-1} x} + C \)
This problem shows how variable substitution can transform a complex integral into a recognizable form.
In simple words: We change the variable from \( x \) to \( t \) by letting \( t = \tan^{-1} x \). This makes the problem look like the special \( e^t \) rule. After finding the answer in terms of \( t \), we change it back to \( x \) to get the final result.

🎯 Exam Tip: For integrals with \( e^{\text{function of x}} \), try substituting \( t \) for the exponent's function. This often simplifies the integral into a solvable form.

Question 6. \( \frac{\log x}{(1 + \log x)^2} \)
Answer: Let the integral be \( I \).
\( I = \int \frac{\log x}{(1 + \log x)^2} \, dx \)
Let's use substitution. Put \( t = \log x \).
This means \( x = e^t \).
Differentiating both sides: \( dx = e^t \, dt \).
Now, substitute these into the integral:
\( \implies I = \int \frac{t}{(1 + t)^2} e^t \, dt \)
We need to manipulate the fraction \( \frac{t}{(1+t)^2} \) to fit the \( f(t) + f'(t) \) form.
We can rewrite \( t \) as \( (1+t) - 1 \):
\( \frac{t}{(1+t)^2} = \frac{(1+t) - 1}{(1+t)^2} \)
Split the fraction:
\( \implies \frac{1+t}{(1+t)^2} - \frac{1}{(1+t)^2} = \frac{1}{1+t} - \frac{1}{(1+t)^2} \)
So the integral becomes:
\( \implies I = \int e^t \left( \frac{1}{1+t} - \frac{1}{(1+t)^2} \right) \, dt \)
This integral is in the form \( \int e^t [f(t) + f'(t)] \, dt \).
Let \( f(t) = \frac{1}{1+t} \).
Then, the derivative \( f'(t) = \frac{d}{dt} \left(\frac{1}{1+t}\right) = -\frac{1}{(1+t)^2} \).
So, we have \( f(t) = \frac{1}{1+t} \) and \( f'(t) = -\frac{1}{(1+t)^2} \).
\( \implies I = e^t \cdot \frac{1}{1+t} + C \)
Now, substitute back \( t = \log x \):
\( \implies I = e^{\log x} \cdot \frac{1}{1+\log x} + C \)
Since \( e^{\log x} = x \):
\( \implies I = x \cdot \frac{1}{1+\log x} + C \)
\( \implies I = \frac{x}{1+\log x} + C \)
This problem shows how adding and subtracting a term in the numerator can help create the required form.
In simple words: First, we use substitution by letting \( t = \log x \). Then, we change the fraction by adding and subtracting 1 in the top part. This helps us split it into two terms where one is the derivative of the other, allowing us to use the special \( e^t \) rule. Finally, we change \( t \) back to \( \log x \).

🎯 Exam Tip: For rational functions of \( \log x \) with \( e^x \), substitution \( t = \log x \) is often the first step. Then, manipulate the fraction by adding/subtracting terms in the numerator to get the \( f(t) + f'(t) \) form.