Samacheer Kalvi Class 12 Business Maths Solutions Chapter 2 Integral Calculus I Exercise 2.4

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.4

Question 1. Integrate the following with respect to x.
\( 2 \cos x - 3 \sin x + 4 \sec^2 x - 5 \operatorname{cosec}^2 x \)
Answer: We need to find the integral of the given expression. We can integrate each term separately.
\[ \int (2 \cos x - 3 \sin x + 4 \sec^2 x - 5 \operatorname{cosec}^2 x) \, dx \] \[ = 2 \int \cos x \, dx - 3 \int \sin x \, dx + 4 \int \sec^2 x \, dx - 5 \int \operatorname{cosec}^2 x \, dx \] \[ = 2 \sin x + 3 \cos x + 4 \tan x + 5 \cot x + c \] The constant 'c' is added because this is an indefinite integral.
In simple words: To integrate this, we break it into smaller parts. We find the integral for each part and then add them up, remembering to put '+ c' at the end.

🎯 Exam Tip: Remember the basic integration formulas for common trigonometric functions. Pay close attention to the signs, especially when integrating \( \sin x \) and \( \operatorname{cosec}^2 x \).

Question 2. Integrate \( \sin^3 x \) with respect to x.
Answer: We use a trigonometric identity to simplify \( \sin^3 x \) before integrating. The identity is:
\( \sin 3x = 3 \sin x - 4 \sin^3 x \)
From this, we can rearrange to find \( \sin^3 x \):
\( 4 \sin^3 x = 3 \sin x - \sin 3x \)
\( \implies \sin^3 x = \frac{1}{4} [3 \sin x - \sin 3x] \) Now we can integrate the simplified expression:
\[ \int \sin^3 x \, dx = \int \frac{1}{4} (3 \sin x - \sin 3x) \, dx \] \[ = \frac{1}{4} \left[ 3 \int \sin x \, dx - \int \sin 3x \, dx \right] \] \[ = \frac{1}{4} \left[ 3(-\cos x) - \left( \frac{-\cos 3x}{3} \right) \right] + c \] \[ = \frac{1}{4} \left[ -3 \cos x + \frac{\cos 3x}{3} \right] + c \] \[ = -\frac{3}{4} \cos x + \frac{1}{12} \cos 3x + c \] This method makes it easier to integrate the cubed sine function.
In simple words: To integrate \( \sin^3 x \), we first change it using a special trig rule to make it simpler. Then, we integrate the new, simpler parts and add them together with '+ c'.

🎯 Exam Tip: When dealing with higher powers of trigonometric functions, look for suitable identities (like for \( \sin^3 x \) or \( \cos^3 x \)) to reduce them to linear terms, making integration much simpler.

Question 3. Integrate \( \frac { \cos 2x+2\sin^2x }{\cos^2x} \) with respect to x.
Answer: First, we use the double angle identity \( \cos 2x = \cos^2 x - \sin^2 x \) to simplify the numerator.
\[ \int \frac{\cos 2x + 2 \sin^2 x}{\cos^2 x} \, dx \] \[ = \int \frac{(\cos^2 x - \sin^2 x) + 2 \sin^2 x}{\cos^2 x} \, dx \] Now, combine the \( \sin^2 x \) terms in the numerator:
\[ = \int \frac{\cos^2 x + \sin^2 x}{\cos^2 x} \, dx \] Since \( \cos^2 x + \sin^2 x = 1 \), the expression simplifies further:
\[ = \int \frac{1}{\cos^2 x} \, dx \] We know that \( \frac{1}{\cos^2 x} \) is the same as \( \sec^2 x \).
\[ = \int \sec^2 x \, dx \] Finally, integrate \( \sec^2 x \):
\[ = \tan x + c \] The trigonometric identities help simplify the expression significantly before integration.
In simple words: First, change the \( \cos 2x \) part using a math rule. Then, simplify the top part of the fraction until it becomes \( \sec^2 x \). After that, we can easily find the integral, which is \( \tan x \) plus a constant.

🎯 Exam Tip: Always look for opportunities to simplify complex expressions using trigonometric identities before attempting integration, as it can often lead to a standard integral form.

Question 4. Integrate \( \frac {1} {\sin^2x \cos^2x} \) with respect to x.
Answer: We can rewrite the numerator using the identity \( \sin^2 x + \cos^2 x = 1 \) to simplify the fraction.
\[ \int \frac{1}{\sin^2 x \cos^2 x} \, dx \] \[ = \int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} \, dx \] Next, split the fraction into two separate terms:
\[ = \int \left( \frac{\sin^2 x}{\sin^2 x \cos^2 x} + \frac{\cos^2 x}{\sin^2 x \cos^2 x} \right) \, dx \] Simplify each term by canceling common factors:
\[ = \int \left( \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} \right) \, dx \] Rewrite these terms using reciprocal identities:
\[ = \int (\sec^2 x + \operatorname{cosec}^2 x) \, dx \] Now, integrate each term separately:
\[ = \int \sec^2 x \, dx + \int \operatorname{cosec}^2 x \, dx \] \[ = \tan x - \cot x + c \] By breaking down the fraction and using identities, we can integrate it easily.
In simple words: We change the number '1' on top to \( \sin^2 x + \cos^2 x \). Then, we split the fraction into two easier parts. These parts become \( \sec^2 x \) and \( \operatorname{cosec}^2 x \), which we can integrate to get \( \tan x - \cot x \).

🎯 Exam Tip: When dealing with fractions involving products of sine and cosine in the denominator, try substituting the numerator with \( \sin^2 x + \cos^2 x \) to separate the terms for easier integration.

Question 5. Integrate \( \sqrt { 1-\sin 2x } \) with respect to x.
Answer: We use trigonometric identities to simplify the expression inside the square root. We know that \( 1 = \sin^2 x + \cos^2 x \) and \( \sin 2x = 2 \sin x \cos x \).
\[ \int \sqrt{1 - \sin 2x} \, dx \] Substitute these identities into the expression:
\[ = \int \sqrt{\sin^2 x + \cos^2 x - 2 \sin x \cos x} \, dx \] The expression inside the square root is a perfect square: \( (a-b)^2 = a^2 - 2ab + b^2 \).
\[ = \int \sqrt{(\sin x - \cos x)^2} \, dx \] Taking the square root, we get:
\[ = \int (\sin x - \cos x) \, dx \] Now, integrate each term separately:
\[ = \int \sin x \, dx - \int \cos x \, dx \] \[ = -\cos x - \sin x + c \] Simplifying the expression under the square root is the key step here.
In simple words: First, change the '1' to \( \sin^2 x + \cos^2 x \) and \( \sin 2x \) to \( 2 \sin x \cos x \). This makes the inside of the square root a perfect square. After taking the square root, we can integrate the simple \( \sin x \) and \( \cos x \) terms.

🎯 Exam Tip: For square root expressions like \( \sqrt{1 \pm \sin 2x} \) or \( \sqrt{1 \pm \cos 2x} \), always try to convert the expression inside the root into a perfect square using relevant trigonometric identities like \( \sin^2 x + \cos^2 x = 1 \) and double angle formulas.