GSEB Class 12 Maths Solutions Chapter 7 Integrals Exercise 7.3

Find the integrals of the following:

Question 1. \( \sin^2(2x + 5) \)
Answer:
Let \( I = \int \sin^2(2x + 5) dx \)
Using the identity \( \sin^2 A = \frac{1 - \cos 2A}{2} \), we get:
\( I = \int \frac{1 - \cos 2(2x + 5)}{2} dx \)
\( I = \frac{1}{2} \int [1 - \cos(4x + 10)] dx \)
\( I = \frac{1}{2} \left[ \int 1 \, dx - \int \cos(4x + 10) dx \right] \)
To integrate \( \cos(4x + 10) \), let \( 4x + 10 = t \). Then \( 4 \, dx = dt \implies dx = \frac{1}{4} dt \).
So, \( \int \cos(4x + 10) dx = \int \cos t \cdot \frac{1}{4} dt = \frac{1}{4} \int \cos t \, dt = \frac{1}{4} \sin t + C_1 = \frac{1}{4} \sin(4x + 10) + C_1 \)
Substituting this back into the expression for \( I \):
\( I = \frac{1}{2} \left[ x - \frac{1}{4} \sin(4x + 10) \right] + C \)
\( I = \frac{x}{2} - \frac{1}{8} \sin(4x + 10) + C \)
In simple words: To solve this, we first change the `sin²` part into a `1-cos` form. Then we integrate each piece separately. For the `cos` part, we use a substitution to simplify it before integrating. Finally, we combine the parts and add the constant `C`.

Exam Tip: Remember to use the double angle formula for `sin^2 x` before integrating, and always handle the coefficient of `x` correctly during substitution.

Question 2. \( \sin 3x \cos 4x \)
Answer:
Let \( I = \int \sin 3x \cos 4x \, dx \)
Using the product-to-sum identity: \( 2 \sin A \cos B = \sin(A+B) + \sin(A-B) \)
So, \( \sin 3x \cos 4x = \frac{1}{2} [\sin(3x + 4x) + \sin(3x - 4x)] \)
\( \sin 3x \cos 4x = \frac{1}{2} [\sin 7x + \sin(-x)] \)
Since \( \sin(-x) = -\sin x \):
\( \sin 3x \cos 4x = \frac{1}{2} [\sin 7x - \sin x] \)
Now, integrate:
\( I = \frac{1}{2} \int [\sin 7x - \sin x] dx \)
\( I = \frac{1}{2} \left[ \int \sin 7x \, dx - \int \sin x \, dx \right] \)
\( I = \frac{1}{2} \left[ \left( -\frac{\cos 7x}{7} \right) - (-\cos x) \right] + C \)
\( I = \frac{1}{2} \left[ -\frac{\cos 7x}{7} + \cos x \right] + C \)
\( I = -\frac{1}{14} \cos 7x + \frac{1}{2} \cos x + C \)
In simple words: We convert the product of `sin` and `cos` into a sum using a special formula. Then, we integrate each `sin` term separately, remembering to adjust for the coefficient of `x`. Finally, we add the constant of integration.

Exam Tip: Always recognize product-to-sum trigonometric identities like `2 sin A cos B` or `2 cos A cos B` and use them to simplify the integral before solving.

Question 3. \( \cos 2x \cos 4x \cos 6x \)
Answer:
Let \( I = \int \cos 2x \cos 4x \cos 6x \, dx \)
First, group two terms and use the identity \( 2 \cos A \cos B = \cos(A+B) + \cos(A-B) \).
Consider \( \cos 2x \cos 4x \):
\( \cos 2x \cos 4x = \frac{1}{2} [2 \cos 4x \cos 2x] \)
\( = \frac{1}{2} [\cos(4x + 2x) + \cos(4x - 2x)] \)
\( = \frac{1}{2} [\cos 6x + \cos 2x] \)
Now substitute this back into the integral:
\( I = \int \frac{1}{2} [\cos 6x + \cos 2x] \cos 6x \, dx \)
\( I = \frac{1}{2} \int [\cos^2 6x + \cos 2x \cos 6x] \, dx \)
Again, use the identities. For \( \cos^2 6x \), use \( \cos^2 A = \frac{1 + \cos 2A}{2} \).
For \( \cos 2x \cos 6x \), use \( 2 \cos A \cos B = \cos(A+B) + \cos(A-B) \).
\( \cos^2 6x = \frac{1 + \cos(2 \cdot 6x)}{2} = \frac{1 + \cos 12x}{2} \)
\( \cos 2x \cos 6x = \frac{1}{2} [\cos(6x + 2x) + \cos(6x - 2x)] = \frac{1}{2} [\cos 8x + \cos 4x] \)
Substitute these into the integral:
\( I = \frac{1}{2} \int \left[ \frac{1 + \cos 12x}{2} + \frac{1}{2} (\cos 8x + \cos 4x) \right] \, dx \)
\( I = \frac{1}{4} \int [1 + \cos 12x + \cos 8x + \cos 4x] \, dx \)
Now integrate each term:
\( I = \frac{1}{4} \left[ x + \frac{\sin 12x}{12} + \frac{\sin 8x}{8} + \frac{\sin 4x}{4} \right] + C \)
\( I = \frac{x}{4} + \frac{\sin 12x}{48} + \frac{\sin 8x}{32} + \frac{\sin 4x}{16} + C \)
In simple words: To solve this, we first change `cos 2x cos 4x` into a sum using a formula. Then we multiply by `cos 6x`. Next, we use two different formulas to change `cos² 6x` and `cos 2x cos 6x` into sums. Finally, we integrate each term in the sum and add the constant `C`.

Exam Tip: When integrating products of three trigonometric functions, apply product-to-sum identities in steps. First, convert two terms, then multiply by the third, and apply identities again if necessary.

Question 4. \( \sin^3(2x + 1) \)
Answer:
Let \( I = \int \sin^3(2x + 1) dx \)
Using the identity \( \sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta \), we can write \( 4 \sin^3 \theta = 3 \sin \theta - \sin 3\theta \), so \( \sin^3 \theta = \frac{1}{4} (3 \sin \theta - \sin 3\theta) \).
Here, \( \theta = 2x + 1 \).
So, \( I = \int \frac{1}{4} [3 \sin(2x + 1) - \sin 3(2x + 1)] dx \)
\( I = \frac{1}{4} \int [3 \sin(2x + 1) - \sin(6x + 3)] dx \)
Now, integrate each term:
\( I = \frac{1}{4} \left[ 3 \int \sin(2x + 1) dx - \int \sin(6x + 3) dx \right] \)
For \( \int \sin(ax+b) dx = -\frac{\cos(ax+b)}{a} + C \):
\( \int \sin(2x + 1) dx = -\frac{\cos(2x + 1)}{2} \)
\( \int \sin(6x + 3) dx = -\frac{\cos(6x + 3)}{6} \)
Substitute these back:
\( I = \frac{1}{4} \left[ 3 \left(-\frac{\cos(2x + 1)}{2}\right) - \left(-\frac{\cos(6x + 3)}{6}\right) \right] + C \)
\( I = \frac{1}{4} \left[ -\frac{3}{2} \cos(2x + 1) + \frac{1}{6} \cos(6x + 3) \right] + C \)
\( I = -\frac{3}{8} \cos(2x + 1) + \frac{1}{24} \cos(6x + 3) + C \)
Alternatively, one can use \( \cos 3\theta = 4 \cos^3 \theta - 3 \cos \theta \) to replace \( \cos(6x+3) \) with \( 4 \cos^3(2x+1) - 3 \cos(2x+1) \), which simplifies the expression to:
\( I = -\frac{3}{8} \cos(2x+1) + \frac{1}{24} [4 \cos^3(2x+1) - 3 \cos(2x+1)] + C \)
\( I = -\frac{3}{8} \cos(2x+1) + \frac{1}{6} \cos^3(2x+1) - \frac{1}{8} \cos(2x+1) + C \)
\( I = \frac{1}{6} \cos^3(2x+1) - \frac{4}{8} \cos(2x+1) + C \)
\( I = \frac{1}{6} \cos^3(2x+1) - \frac{1}{2} \cos(2x+1) + C \)
In simple words: We use a trigonometric identity for `sin³θ` to convert it into a form that is easier to integrate. Then we integrate each `sin` term, keeping track of the coefficients for `x`. This gives us the final answer.

Exam Tip: When dealing with powers of `sin` or `cos`, always remember to use the identities `sin^3 θ = (3 sin θ - sin 3θ)/4` or `cos^3 θ = (3 cos θ + cos 3θ)/4` to simplify the expression into single powers of `sin` or `cos` before integrating.

Question 5. \( \sin^3 x \cos^3 x \)
Answer:
Let \( I = \int \sin^3 x \cos^3 x \, dx \)
We can rewrite this as: \( I = \int (\sin x \cos x)^3 dx \)
Using the identity \( \sin 2x = 2 \sin x \cos x \), so \( \sin x \cos x = \frac{\sin 2x}{2} \).
\( I = \int \left(\frac{\sin 2x}{2}\right)^3 dx = \int \frac{\sin^3 2x}{8} dx = \frac{1}{8} \int \sin^3 2x \, dx \)
Now, use the identity \( \sin^3 \theta = \frac{1}{4} (3 \sin \theta - \sin 3\theta) \), where \( \theta = 2x \):
\( I = \frac{1}{8} \int \frac{1}{4} [3 \sin 2x - \sin(3 \cdot 2x)] dx \)
\( I = \frac{1}{32} \int [3 \sin 2x - \sin 6x] dx \)
Integrate each term:
\( I = \frac{1}{32} \left[ 3 \left(-\frac{\cos 2x}{2}\right) - \left(-\frac{\cos 6x}{6}\right) \right] + C \)
\( I = \frac{1}{32} \left[ -\frac{3}{2} \cos 2x + \frac{1}{6} \cos 6x \right] + C \)
\( I = -\frac{3}{64} \cos 2x + \frac{1}{192} \cos 6x + C \)
Alternatively, we can express one power in terms of the other:
\( I = \int \sin^3 x \cos^3 x \, dx = \int \sin^3 x \cos^2 x \cos x \, dx \)
\( = \int \sin^3 x (1 - \sin^2 x) \cos x \, dx \)
Let \( u = \sin x \). Then \( du = \cos x \, dx \).
\( I = \int u^3 (1 - u^2) du = \int (u^3 - u^5) du \)
\( I = \frac{u^4}{4} - \frac{u^6}{6} + C \)
Substitute back \( u = \sin x \):
\( I = \frac{\sin^4 x}{4} - \frac{\sin^6 x}{6} + C \)
The solution in the OCR uses a different approach: \( I = \int \sin x \cdot \sin^2 x \cdot \cos^3 x \, dx = \int \sin x (1 - \cos^2 x) \cos^3 x \, dx \)
\( = \int (\cos^3 x - \cos^5 x) \sin x \, dx \)
Let \( t = \cos x \). Then \( dt = -\sin x \, dx \implies \sin x \, dx = -dt \).
\( I = \int (t^3 - t^5) (-dt) = \int (t^5 - t^3) dt \)
\( I = \frac{t^6}{6} - \frac{t^4}{4} + C \)
Substitute back \( t = \cos x \):
\( I = \frac{\cos^6 x}{6} - \frac{\cos^4 x}{4} + C \)
In simple words: We rewrite the expression by splitting `sin³x` or `cos³x` and using the `sin²x + cos²x = 1` identity. Then, we use a substitution like `t = cos x` or `t = sin x` to simplify the integral. Finally, we integrate the resulting polynomial in `t` and substitute back to get the answer.

Exam Tip: When integrating products of odd powers of `sin` and `cos`, pull out one `sin x` or `cos x` term, convert the remaining even power using `sin^2 x = 1 - cos^2 x` or `cos^2 x = 1 - sin^2 x`, and then use substitution (`u = cos x` or `u = sin x`).

Question 6. \( \sin x \sin 2x \sin 3x \)
Answer:
Let \( I = \int \sin x \sin 2x \sin 3x \, dx \)
First, apply the product-to-sum identity to \( \sin x \sin 2x \): \( 2 \sin A \sin B = \cos(A-B) - \cos(A+B) \).
\( \sin x \sin 2x = \frac{1}{2} [\cos(x - 2x) - \cos(x + 2x)] \)
\( = \frac{1}{2} [\cos(-x) - \cos 3x] \)
Since \( \cos(-x) = \cos x \):
\( = \frac{1}{2} [\cos x - \cos 3x] \)
Now substitute this back into the integral:
\( I = \int \frac{1}{2} [\cos x - \cos 3x] \sin 3x \, dx \)
\( I = \frac{1}{2} \int [\cos x \sin 3x - \cos 3x \sin 3x] \, dx \)
Apply product-to-sum identities again for both terms:
For \( \cos x \sin 3x \): \( 2 \cos A \sin B = \sin(A+B) - \sin(A-B) \)
\( \cos x \sin 3x = \frac{1}{2} [\sin(x + 3x) - \sin(x - 3x)] \)
\( = \frac{1}{2} [\sin 4x - \sin(-2x)] \)
\( = \frac{1}{2} [\sin 4x + \sin 2x] \)
For \( \cos 3x \sin 3x \): \( \sin 2A = 2 \sin A \cos A \implies \sin A \cos A = \frac{\sin 2A}{2} \)
\( \cos 3x \sin 3x = \frac{\sin(2 \cdot 3x)}{2} = \frac{\sin 6x}{2} \)
Substitute these back into \( I \):
\( I = \frac{1}{2} \int \left[ \frac{1}{2}(\sin 4x + \sin 2x) - \frac{\sin 6x}{2} \right] \, dx \)
\( I = \frac{1}{4} \int [\sin 4x + \sin 2x - \sin 6x] \, dx \)
Now, integrate each term:
\( I = \frac{1}{4} \left[ -\frac{\cos 4x}{4} - \frac{\cos 2x}{2} - \left(-\frac{\cos 6x}{6}\right) \right] + C \)
\( I = \frac{1}{4} \left[ -\frac{\cos 4x}{4} - \frac{\cos 2x}{2} + \frac{\cos 6x}{6} \right] + C \)
\( I = -\frac{1}{16} \cos 4x - \frac{1}{8} \cos 2x + \frac{1}{24} \cos 6x + C \)
In simple words: To integrate this product of three sine terms, we apply trigonometric formulas twice. First, we convert `sin x sin 2x` into a sum or difference of `cos` terms. Then, we multiply the result by `sin 3x` and apply more formulas to get a sum of simple `sin` terms. Finally, we integrate each term.

Exam Tip: When dealing with products of three trigonometric functions, simplify in stages using appropriate product-to-sum identities. Start with any two terms, then integrate the result with the third.

Question 7. \( \sin 4x \sin 8x \)
Answer:
Let \( I = \int \sin 4x \sin 8x \, dx \)
Using the product-to-sum identity: \( 2 \sin A \sin B = \cos(A-B) - \cos(A+B) \)
So, \( \sin 4x \sin 8x = \frac{1}{2} [2 \sin 4x \sin 8x] \)
\( = \frac{1}{2} [\cos(4x - 8x) - \cos(4x + 8x)] \)
\( = \frac{1}{2} [\cos(-4x) - \cos 12x] \)
Since \( \cos(-4x) = \cos 4x \):
\( = \frac{1}{2} [\cos 4x - \cos 12x] \)
Now, integrate:
\( I = \frac{1}{2} \int [\cos 4x - \cos 12x] dx \)
\( I = \frac{1}{2} \left[ \int \cos 4x \, dx - \int \cos 12x \, dx \right] \)
\( I = \frac{1}{2} \left[ \frac{\sin 4x}{4} - \frac{\sin 12x}{12} \right] + C \)
\( I = \frac{\sin 4x}{8} - \frac{\sin 12x}{24} + C \)
In simple words: We convert the product of two `sin` terms into a difference of `cos` terms using a trigonometric formula. Then, we integrate each `cos` term separately, adjusting for the coefficient of `x`. Finally, we add the constant `C`.

Exam Tip: For products of `sin` functions, use the identity `2 sin A sin B = cos(A-B) - cos(A+B)`. This simplifies the product into terms that are easy to integrate directly.

Question 8. \( \frac{1-\cos x}{1+\cos x} \)
Answer:
Let \( I = \int \frac{1-\cos x}{1+\cos x} \, dx \)
Using the half-angle identities: \( 1 - \cos x = 2 \sin^2 \frac{x}{2} \) and \( 1 + \cos x = 2 \cos^2 \frac{x}{2} \).
Substitute these into the integral:
\( I = \int \frac{2 \sin^2 \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \, dx \)
\( I = \int \frac{\sin^2 \frac{x}{2}}{\cos^2 \frac{x}{2}} \, dx \)
\( I = \int \tan^2 \frac{x}{2} \, dx \)
Using the identity \( \tan^2 \theta = \sec^2 \theta - 1 \):
\( I = \int \left( \sec^2 \frac{x}{2} - 1 \right) dx \)
Now integrate each term:
\( I = \int \sec^2 \frac{x}{2} \, dx - \int 1 \, dx \)
For \( \int \sec^2(ax+b) dx = \frac{\tan(ax+b)}{a} + C \):
\( \int \sec^2 \frac{x}{2} \, dx = \frac{\tan \frac{x}{2}}{\frac{1}{2}} = 2 \tan \frac{x}{2} \)
So, \( I = 2 \tan \frac{x}{2} - x + C \)
In simple words: We simplify the fraction using half-angle trigonometric formulas for `1-cos x` and `1+cos x`. This changes the expression into `tan²(x/2)`. Then, we use another identity to write `tan²(x/2)` as `sec²(x/2) - 1`, which we can easily integrate term by term to get the final answer.

Exam Tip: Whenever you see `(1 - cos x) / (1 + cos x)` or similar forms, immediately think of half-angle formulas to convert them into `tan^2(x/2)` which is easily integrable as `sec^2(x/2) - 1`.

Question 9. \( \frac{\cos x}{1+\cos x} \)
Answer:
Let \( I = \int \frac{\cos x}{1+\cos x} \, dx \)
We can rewrite the numerator to separate the terms:
\( I = \int \frac{(1 + \cos x) - 1}{1+\cos x} \, dx \)
\( I = \int \left( \frac{1+\cos x}{1+\cos x} - \frac{1}{1+\cos x} \right) dx \)
\( I = \int \left( 1 - \frac{1}{1+\cos x} \right) dx \)
Now, integrate each term. For \( \frac{1}{1+\cos x} \), use the half-angle identity \( 1 + \cos x = 2 \cos^2 \frac{x}{2} \).
\( I = \int 1 \, dx - \int \frac{1}{2 \cos^2 \frac{x}{2}} \, dx \)
\( I = x - \frac{1}{2} \int \frac{1}{\cos^2 \frac{x}{2}} \, dx \)
\( I = x - \frac{1}{2} \int \sec^2 \frac{x}{2} \, dx \)
For \( \int \sec^2(ax+b) dx = \frac{\tan(ax+b)}{a} + C \):
\( \int \sec^2 \frac{x}{2} \, dx = \frac{\tan \frac{x}{2}}{\frac{1}{2}} = 2 \tan \frac{x}{2} \)
So, \( I = x - \frac{1}{2} \left( 2 \tan \frac{x}{2} \right) + C \)
\( I = x - \tan \frac{x}{2} + C \)
In simple words: To solve this, we rewrite the numerator by adding and subtracting 1, allowing us to split the fraction. Then, we use a half-angle identity for `1+cos x` to simplify the second part to `sec²(x/2)`. Finally, we integrate each part, which involves simple integration rules for `x` and `tan(x/2)`.

Exam Tip: When the numerator is `cos x` and the denominator is `1 + cos x`, add and subtract `1` in the numerator. This helps split the fraction into `1` and `1/(1+cos x)`, which are easier to integrate using half-angle formulas.

Question 10. \( \sin^4 x \)
Answer:
Let \( I = \int \sin^4 x \, dx \)
We can rewrite \( \sin^4 x \) as \( (\sin^2 x)^2 \).
Using the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \):
\( I = \int \left( \frac{1 - \cos 2x}{2} \right)^2 dx \)
\( I = \int \frac{1 - 2 \cos 2x + \cos^2 2x}{4} \, dx \)
\( I = \frac{1}{4} \int (1 - 2 \cos 2x + \cos^2 2x) \, dx \)
Now, use the identity \( \cos^2 A = \frac{1 + \cos 2A}{2} \) for \( \cos^2 2x \):
\( \cos^2 2x = \frac{1 + \cos(2 \cdot 2x)}{2} = \frac{1 + \cos 4x}{2} \)
Substitute this back into the integral:
\( I = \frac{1}{4} \int \left( 1 - 2 \cos 2x + \frac{1 + \cos 4x}{2} \right) dx \)
\( I = \frac{1}{4} \int \left( 1 - 2 \cos 2x + \frac{1}{2} + \frac{\cos 4x}{2} \right) dx \)
Combine the constant terms:
\( I = \frac{1}{4} \int \left( \frac{3}{2} - 2 \cos 2x + \frac{1}{2} \cos 4x \right) dx \)
Now, integrate each term:
\( I = \frac{1}{4} \left[ \frac{3}{2} x - 2 \left(\frac{\sin 2x}{2}\right) + \frac{1}{2} \left(\frac{\sin 4x}{4}\right) \right] + C \)
\( I = \frac{1}{4} \left[ \frac{3}{2} x - \sin 2x + \frac{\sin 4x}{8} \right] + C \)
\( I = \frac{3}{8} x - \frac{1}{4} \sin 2x + \frac{1}{32} \sin 4x + C \)
In simple words: To integrate `sin⁴x`, we rewrite it as `(sin²x)²`. Then we use the double angle formula to change `sin²x` into `(1-cos 2x)/2`. After squaring and expanding, we apply the `cos²A` formula again for the `cos²2x` term. Finally, we integrate each simplified `cos` and constant term individually.

Exam Tip: For even powers of `sin` or `cos`, repeatedly use the half-angle identities `sin^2 x = (1 - cos 2x)/2` or `cos^2 x = (1 + cos 2x)/2` until all terms are in linear powers of `cos` (or `sin`), which are easy to integrate.

Question 11. \( \cos^4(2x) \)
Answer:
Let \( I = \int \cos^4(2x) \, dx \)
We can rewrite \( \cos^4(2x) \) as \( (\cos^2(2x))^2 \).
Using the identity \( \cos^2 A = \frac{1 + \cos 2A}{2} \):
\( \cos^2(2x) = \frac{1 + \cos(2 \cdot 2x)}{2} = \frac{1 + \cos 4x}{2} \)
Substitute this into the integral:
\( I = \int \left( \frac{1 + \cos 4x}{2} \right)^2 dx \)
\( I = \int \frac{1 + 2 \cos 4x + \cos^2 4x}{4} \, dx \)
\( I = \frac{1}{4} \int (1 + 2 \cos 4x + \cos^2 4x) \, dx \)
Now, use the identity \( \cos^2 A = \frac{1 + \cos 2A}{2} \) for \( \cos^2 4x \):
\( \cos^2 4x = \frac{1 + \cos(2 \cdot 4x)}{2} = \frac{1 + \cos 8x}{2} \)
Substitute this back into the integral:
\( I = \frac{1}{4} \int \left( 1 + 2 \cos 4x + \frac{1 + \cos 8x}{2} \right) dx \)
\( I = \frac{1}{4} \int \left( 1 + 2 \cos 4x + \frac{1}{2} + \frac{\cos 8x}{2} \right) dx \)
Combine the constant terms:
\( I = \frac{1}{4} \int \left( \frac{3}{2} + 2 \cos 4x + \frac{1}{2} \cos 8x \right) dx \)
Now, integrate each term:
\( I = \frac{1}{4} \left[ \frac{3}{2} x + 2 \left(\frac{\sin 4x}{4}\right) + \frac{1}{2} \left(\frac{\sin 8x}{8}\right) \right] + C \)
\( I = \frac{1}{4} \left[ \frac{3}{2} x + \frac{\sin 4x}{2} + \frac{\sin 8x}{16} \right] + C \)
\( I = \frac{3}{8} x + \frac{1}{8} \sin 4x + \frac{1}{64} \sin 8x + C \)
In simple words: To integrate `cos⁴(2x)`, we first write it as `(cos²(2x))²`. Then, we apply the double angle formula for `cos²A` to `cos²(2x)`. After expanding the square, we use the `cos²A` formula again for the `cos²4x` term. Finally, we integrate each simplified constant and `cos` term.

Exam Tip: Remember to carry the argument `(2x)` correctly through all trigonometric identities. For `cos^4(2x)`, the first reduction uses `2A = 2(2x) = 4x`, and the second uses `2A = 2(4x) = 8x`.

Question 12. \( \frac{\sin^2 x}{1+\cos x} \)
Answer:
Let \( I = \int \frac{\sin^2 x}{1+\cos x} \, dx \)
Using the identity \( \sin^2 x = 1 - \cos^2 x \):
\( I = \int \frac{1 - \cos^2 x}{1+\cos x} \, dx \)
Factor the numerator using \( a^2 - b^2 = (a-b)(a+b) \):
\( I = \int \frac{(1 - \cos x)(1 + \cos x)}{1+\cos x} \, dx \)
Cancel out the common term \( (1 + \cos x) \) (assuming \( 1+\cos x \neq 0 \)):
\( I = \int (1 - \cos x) \, dx \)
Now, integrate each term:
\( I = \int 1 \, dx - \int \cos x \, dx \)
\( I = x - \sin x + C \)
In simple words: We start by replacing `sin²x` with `1 - cos²x` using a basic trigonometric identity. Then, we factor the numerator and cancel out the `(1 + cos x)` term from both the top and bottom. This leaves us with a simple `1 - cos x` expression, which is straightforward to integrate.

Exam Tip: When you see `sin^2 x` in the numerator with a `(1 + cos x)` or `(1 - cos x)` in the denominator, always try using the identity `sin^2 x = 1 - cos^2 x` and factoring the numerator to simplify the fraction before integrating.

Question 13. \( \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} \)
Answer:
Let \( I = \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} \, dx \)
Using the double angle identity \( \cos 2\theta = 2 \cos^2 \theta - 1 \):
\( I = \int \frac{(2 \cos^2 x - 1) - (2 \cos^2 \alpha - 1)}{\cos x - \cos \alpha} \, dx \)
\( I = \int \frac{2 \cos^2 x - 1 - 2 \cos^2 \alpha + 1}{\cos x - \cos \alpha} \, dx \)
\( I = \int \frac{2 \cos^2 x - 2 \cos^2 \alpha}{\cos x - \cos \alpha} \, dx \)
Factor out 2 from the numerator:
\( I = \int \frac{2 (\cos^2 x - \cos^2 \alpha)}{\cos x - \cos \alpha} \, dx \)
Factor the numerator using \( a^2 - b^2 = (a-b)(a+b) \):
\( I = \int \frac{2 (\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha} \, dx \)
Cancel out the common term \( (\cos x - \cos \alpha) \) (assuming \( \cos x - \cos \alpha \neq 0 \)):
\( I = \int 2 (\cos x + \cos \alpha) \, dx \)
Since \( \alpha \) is a constant, \( \cos \alpha \) is also a constant.
Integrate each term:
\( I = 2 \int \cos x \, dx + 2 \int \cos \alpha \, dx \)
\( I = 2 \sin x + 2 (\cos \alpha) x + C \)
\( I = 2 (\sin x + x \cos \alpha) + C \)
In simple words: We simplify the fraction using the double angle formula for `cos 2x` and `cos 2α`. This allows us to factor the numerator as a difference of squares. After canceling the common term with the denominator, we are left with a simple expression `2(cos x + cos α)`, which we integrate term by term. Since `α` is a constant, `cos α` is also a constant.

Exam Tip: When you see `cos 2x` and `cos 2α`, recall the identity `cos 2θ = 2 cos^2 θ - 1`. This allows you to convert the expression into terms of `cos x` and `cos α`, making it factorable.

Question 14. \( \frac{\cos x - \sin x}{1 + \sin 2x} \)
Answer:
Let \( I = \int \frac{\cos x - \sin x}{1 + \sin 2x} \, dx \)
Recall that \( 1 = \sin^2 x + \cos^2 x \) and \( \sin 2x = 2 \sin x \cos x \).
So, the denominator \( 1 + \sin 2x \) can be rewritten as:
\( 1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x = (\sin x + \cos x)^2 \)
Substitute this into the integral:
\( I = \int \frac{\cos x - \sin x}{(\sin x + \cos x)^2} \, dx \)
Now, use substitution. Let \( t = \sin x + \cos x \).
Then \( dt = (\cos x - \sin x) \, dx \).
Substitute \( t \) and \( dt \) into the integral:
\( I = \int \frac{dt}{t^2} \)
\( I = \int t^{-2} dt \)
Integrate using the power rule \( \int u^n du = \frac{u^{n+1}}{n+1} + C \):
\( I = \frac{t^{-2+1}}{-2+1} + C = \frac{t^{-1}}{-1} + C = -\frac{1}{t} + C \)
Substitute back \( t = \sin x + \cos x \):
\( I = -\frac{1}{\sin x + \cos x} + C \)
In simple words: We simplify the denominator `1 + sin 2x` by recognizing it as `(sin x + cos x)²`. Then, we use a substitution where `t` is `sin x + cos x`, and `dt` becomes the numerator `(cos x - sin x) dx`. This changes the integral into a simple power form of `t`, which we integrate easily and then substitute `t` back to get the final answer.

Exam Tip: When you see `1 + sin 2x` or `1 - sin 2x` in the denominator with `(cos x - sin x)` or `(cos x + sin x)` in the numerator, recognize that `1 ± sin 2x` can be rewritten as `(sin x ± cos x)^2`. This is a strong hint for a substitution where `t = sin x ± cos x`.

Question 15. \( \tan^3 2x \sec 2x \)
Answer:
Let \( I = \int \tan^3 2x \sec 2x \, dx \)
We can rewrite \( \tan^3 2x \sec 2x \) as \( \tan^2 2x \cdot (\tan 2x \sec 2x) \).
Using the identity \( \tan^2 \theta = \sec^2 \theta - 1 \):
\( I = \int (\sec^2 2x - 1) (\tan 2x \sec 2x) \, dx \)
Now, use substitution. Let \( t = \sec 2x \).
Then \( dt = \frac{d}{dx}(\sec 2x) \, dx = (\sec 2x \tan 2x \cdot 2) \, dx = 2 \sec 2x \tan 2x \, dx \).
So, \( \tan 2x \sec 2x \, dx = \frac{1}{2} dt \).
Substitute \( t \) and \( dt \) into the integral:
\( I = \int (t^2 - 1) \frac{1}{2} dt \)
\( I = \frac{1}{2} \int (t^2 - 1) dt \)
Integrate each term:
\( I = \frac{1}{2} \left( \frac{t^3}{3} - t \right) + C \)
Substitute back \( t = \sec 2x \):
\( I = \frac{1}{2} \left( \frac{\sec^3 2x}{3} - \sec 2x \right) + C \)
\( I = \frac{1}{6} \sec^3 2x - \frac{1}{2} \sec 2x + C \)
In simple words: We rewrite the `tan³2x` part as `tan²2x` times `tan 2x`. Then, we use the identity to change `tan²2x` to `sec²2x - 1`. This makes the integral ready for substitution. We let `t` be `sec 2x`, which means `dt` includes the `tan 2x sec 2x` part. After substitution, we integrate the polynomial in `t` and then replace `t` with `sec 2x` to get the final answer.

Exam Tip: For integrals of the form `tan^m x sec^n x` where `m` is odd and `n >= 1`, pull out `sec x tan x` and convert the remaining `tan^2 x` terms to `sec^2 x - 1`. Then use the substitution `u = sec x`.

Question 16. \( \tan^4 x \)
Answer:
Let \( I = \int \tan^4 x \, dx \)
We can rewrite \( \tan^4 x \) as \( \tan^2 x \cdot \tan^2 x \).
Using the identity \( \tan^2 x = \sec^2 x - 1 \):
\( I = \int \tan^2 x (\sec^2 x - 1) \, dx \)
Distribute \( \tan^2 x \):
\( I = \int (\tan^2 x \sec^2 x - \tan^2 x) \, dx \)
\( I = \int \tan^2 x \sec^2 x \, dx - \int \tan^2 x \, dx \)
For the first integral, \( \int \tan^2 x \sec^2 x \, dx \), let \( t = \tan x \). Then \( dt = \sec^2 x \, dx \).
So, \( \int \tan^2 x \sec^2 x \, dx = \int t^2 dt = \frac{t^3}{3} + C_1 = \frac{\tan^3 x}{3} + C_1 \).
For the second integral, \( \int \tan^2 x \, dx \), use the identity \( \tan^2 x = \sec^2 x - 1 \):
\( \int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx = \int \sec^2 x \, dx - \int 1 \, dx = \tan x - x + C_2 \).
Combine the results:
\( I = \frac{\tan^3 x}{3} - (\tan x - x) + C \)
\( I = \frac{1}{3} \tan^3 x - \tan x + x + C \)
In simple words: To integrate `tan⁴x`, we break it into `tan²x` times `tan²x`. Then, we change one `tan²x` to `sec²x - 1`. This creates two separate integrals: one that can be solved with a simple substitution (`t = tan x`) and another that involves integrating `sec²x - 1`. We combine the results for the final answer.

Exam Tip: When integrating even powers of `tan x`, pull out `tan^2 x` and convert it to `sec^2 x - 1`. This usually results in terms that are either directly integrable or solvable by simple substitution.

Question 17. \( \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} \)
Answer:
Let \( I = \int \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} \, dx \)
Split the fraction into two separate terms:
\( I = \int \left( \frac{\sin^3 x}{\sin^2 x \cos^2 x} + \frac{\cos^3 x}{\sin^2 x \cos^2 x} \right) dx \)
Simplify each term:
\( I = \int \left( \frac{\sin x}{\cos^2 x} + \frac{\cos x}{\sin^2 x} \right) dx \)
Rewrite the terms using `sec x`, `tan x`, `cosec x`, `cot x`:
\( \frac{\sin x}{\cos^2 x} = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = \tan x \sec x \)
\( \frac{\cos x}{\sin^2 x} = \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} = \cot x \cosec x \)
So, \( I = \int (\tan x \sec x + \cot x \cosec x) \, dx \)
Integrate each term:
\( I = \int \tan x \sec x \, dx + \int \cot x \cosec x \, dx \)
Recall that \( \int \sec x \tan x \, dx = \sec x \) and \( \int \cosec x \cot x \, dx = -\cosec x \).
\( I = \sec x - \cosec x + C \)
In simple words: We first divide the fraction into two parts, separating the `sin³x` and `cos³x` terms. Then, we simplify each part by canceling common `sin` and `cos` terms, which transforms them into `sec x tan x` and `cosec x cot x`. Finally, we integrate these standard trigonometric forms to get the answer.

Exam Tip: When faced with a sum of trigonometric powers in the numerator and a product in the denominator, split the fraction. This often leads to terms that simplify into standard integral forms like `sec x tan x` or `cosec x cot x`.

Question 18. \( \frac{\cos 2x + 2 \sin^2 x}{\cos^2 x} \)
Answer:
Let \( I = \int \frac{\cos 2x + 2 \sin^2 x}{\cos^2 x} \, dx \)
Using the double angle identity \( \cos 2x = \cos^2 x - \sin^2 x \):
\( I = \int \frac{(\cos^2 x - \sin^2 x) + 2 \sin^2 x}{\cos^2 x} \, dx \)
Combine the \( \sin^2 x \) terms in the numerator:
\( I = \int \frac{\cos^2 x + \sin^2 x}{\cos^2 x} \, dx \)
Using the identity \( \sin^2 x + \cos^2 x = 1 \):
\( I = \int \frac{1}{\cos^2 x} \, dx \)
Rewrite in terms of `sec x`:
\( I = \int \sec^2 x \, dx \)
Integrate \( \sec^2 x \):
\( I = \tan x + C \)
In simple words: We simplify the numerator using the double angle formula for `cos 2x`. Then, we combine the `sin²x` terms in the numerator. The numerator then becomes `cos²x + sin²x`, which simplifies to `1`. So, the integral reduces to `1/cos²x`, which is `sec²x`. We then integrate `sec²x` to get `tan x`.

Exam Tip: When `cos 2x` is present in an integrand, consider using the identity `cos 2x = cos^2 x - sin^2 x` (or its variations) to simplify it, especially if other `sin^2 x` or `cos^2 x` terms are present. This often leads to significant cancellation.

Question 19. \( \frac{1}{\sin x \cos^3 x} \)
Answer:
Let \( I = \int \frac{1}{\sin x \cos^3 x} \, dx \)
We can rewrite `1` as `sin²x + cos²x` in the numerator:
\( I = \int \frac{\sin^2 x + \cos^2 x}{\sin x \cos^3 x} \, dx \)
Split the fraction into two separate terms:
\( I = \int \left( \frac{\sin^2 x}{\sin x \cos^3 x} + \frac{\cos^2 x}{\sin x \cos^3 x} \right) dx \)
Simplify each term:
\( I = \int \left( \frac{\sin x}{\cos^3 x} + \frac{1}{\sin x \cos x} \right) dx \)
Rewrite the terms: For \( \frac{\sin x}{\cos^3 x} = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos^2 x} = \tan x \sec^2 x \).
For \( \frac{1}{\sin x \cos x} \), multiply numerator and denominator by `2` to get `sin 2x`:
\( \frac{1}{\sin x \cos x} = \frac{2}{2 \sin x \cos x} = \frac{2}{\sin 2x} = 2 \cosec 2x \).
So, \( I = \int (\tan x \sec^2 x + 2 \cosec 2x) \, dx \)
Integrate each term:
\( I = \int \tan x \sec^2 x \, dx + 2 \int \cosec 2x \, dx \)
For \( \int \tan x \sec^2 x \, dx \), let \( t = \tan x \). Then \( dt = \sec^2 x \, dx \).
So, \( \int \tan x \sec^2 x \, dx = \int t \, dt = \frac{t^2}{2} + C_1 = \frac{\tan^2 x}{2} + C_1 \).
For \( \int \cosec 2x \, dx \), recall \( \int \cosec \theta \, d\theta = \ln |\tan(\theta/2)| \).
So, \( \int \cosec 2x \, dx = \frac{1}{2} \ln |\tan x| + C_2 \).
Therefore, \( I = \frac{\tan^2 x}{2} + 2 \left( \frac{1}{2} \ln |\tan x| \right) + C \)
\( I = \frac{1}{2} \tan^2 x + \ln |\tan x| + C \)
The OCR solution provides an alternative simplification for the second term:
\( \frac{1}{\sin x \cos x} \cdot \frac{\sec^2 x}{\sec^2 x} = \frac{\sec^2 x}{\tan x} \).
So, \( I = \int \left( \tan x \sec^2 x + \frac{\sec^2 x}{\tan x} \right) dx \)
Let \( t = \tan x \). Then \( dt = \sec^2 x \, dx \).
\( I = \int \left( t + \frac{1}{t} \right) dt \)
\( I = \int t \, dt + \int \frac{1}{t} \, dt \)
\( I = \frac{t^2}{2} + \ln |t| + C \)
Substitute back \( t = \tan x \):
\( I = \frac{\tan^2 x}{2} + \ln |\tan x| + C \)
In simple words: We rewrite `1` in the numerator as `sin²x + cos²x` and then split the fraction. Each resulting term is simplified using trigonometric identities. One term becomes `tan x sec²x`, which is integrated using substitution. The other term becomes `sec²x/tan x`, which also simplifies with substitution. Finally, we combine these integrated parts for the complete answer.

Exam Tip: When the numerator is `1` and the denominator has products of `sin x` and `cos x` with odd powers (like `cos^3 x`), try replacing `1` with `sin^2 x + cos^2 x` and splitting the fraction. This often creates terms that are directly integrable or suitable for `u = tan x` substitution.

Question 20. \( \frac{\cos 2x}{(\cos x + \sin x)^2} \)
Answer:
Let \( I = \int \frac{\cos 2x}{(\cos x + \sin x)^2} \, dx \)
Using the double angle identity \( \cos 2x = \cos^2 x - \sin^2 x \):
\( I = \int \frac{\cos^2 x - \sin^2 x}{(\cos x + \sin x)^2} \, dx \)
Factor the numerator using \( a^2 - b^2 = (a-b)(a+b) \):
\( I = \int \frac{(\cos x - \sin x)(\cos x + \sin x)}{(\cos x + \sin x)^2} \, dx \)
Cancel out the common term \( (\cos x + \sin x) \) (assuming \( \cos x + \sin x \neq 0 \)):
\( I = \int \frac{\cos x - \sin x}{\cos x + \sin x} \, dx \)
Now, use substitution. Let \( t = \cos x + \sin x \).
Then \( dt = (-\sin x + \cos x) \, dx = (\cos x - \sin x) \, dx \).
Substitute \( t \) and \( dt \) into the integral:
\( I = \int \frac{dt}{t} \)
Integrate \( \frac{1}{t} \):
\( I = \ln |t| + C \)
Substitute back \( t = \cos x + \sin x \):
\( I = \ln |\cos x + \sin x| + C \)
In simple words: We rewrite `cos 2x` in the numerator as `cos²x - sin²x` using a double angle identity. Then we factor the numerator as a difference of squares and cancel a common term with the denominator. This simplifies the expression to `(cos x - sin x) / (cos x + sin x)`. We then use a substitution where `t` is `cos x + sin x`, and `dt` becomes the numerator. The integral becomes `1/t dt`, which integrates to `ln|t|`. Finally, we substitute `t` back to get the result.

Exam Tip: When the numerator is `cos 2x` and the denominator is `(cos x ± sin x)^2`, always try using the identity `cos 2x = cos^2 x - sin^2 x` to factor the numerator, leading to easy cancellation and a simple logarithmic integral via substitution.

Question 21. \( \sin^{-1} x \)
Answer:
Let \( I = \int \sin^{-1} x \, dx \)
This integral requires integration by parts. Let \( u = \sin^{-1} x \) and \( dv = dx \).
Then \( du = \frac{1}{\sqrt{1 - x^2}} dx \) and \( v = x \).
The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \).
\( I = x \sin^{-1} x - \int x \cdot \frac{1}{\sqrt{1 - x^2}} \, dx \)
Now, we need to solve \( \int \frac{x}{\sqrt{1 - x^2}} \, dx \).
Let \( w = 1 - x^2 \). Then \( dw = -2x \, dx \implies x \, dx = -\frac{1}{2} dw \).
So, \( \int \frac{x}{\sqrt{1 - x^2}} \, dx = \int \frac{-\frac{1}{2} dw}{\sqrt{w}} = -\frac{1}{2} \int w^{-1/2} dw \)
\( = -\frac{1}{2} \frac{w^{1/2}}{1/2} + C' = -w^{1/2} + C' = -\sqrt{1 - x^2} + C' \)
Substitute this back into the expression for \( I \):
\( I = x \sin^{-1} x - (-\sqrt{1 - x^2}) + C \)
\( I = x \sin^{-1} x + \sqrt{1 - x^2} + C \)
The OCR solution shows `sin⁻¹(cos x)`. Let's address that alternative if the question was `sin⁻¹(cos x)`.
If \( I = \int \sin^{-1}(\cos x) dx \)
We know that \( \sin^{-1}(\cos x) = \sin^{-1}(\sin(\frac{\pi}{2} - x)) = \frac{\pi}{2} - x \) for \( x \in [0, \pi] \).
So, \( I = \int \left(\frac{\pi}{2} - x\right) dx \)
\( I = \int \frac{\pi}{2} dx - \int x \, dx \)
\( I = \frac{\pi}{2} x - \frac{x^2}{2} + C \)
Given the OCR question just said `sin⁻¹x`, the first solution is correct. The OCR solution follows the second interpretation by using the identity. The OCR text "Solution: Let I = ∫sin⁻¹(cosx)dx = ∫sin⁻¹[sin(π/2 - x)]dx = ∫(π/2-x)dx = π/2∫dx - ∫xdx = πx/2 - x²/2 + C." is a solution to `sin⁻¹(cos x)`, not `sin⁻¹x`. I will provide the answer for the question `sin⁻1x` and then the OCR solution for `sin⁻1(cos x)`.

**Answer based on OCR's interpretation for `sin⁻¹(cos x)`:**
Let \( I = \int \sin^{-1}(\cos x) dx \)
Using the identity \( \sin^{-1}(\cos x) = \frac{\pi}{2} - x \) (for appropriate range of x)
\( I = \int \left(\frac{\pi}{2} - x\right) dx \)
\( I = \int \frac{\pi}{2} dx - \int x \, dx \)
\( I = \frac{\pi}{2} x - \frac{x^2}{2} + C \)
In simple words: If the question is `sin⁻¹x`, we use integration by parts, differentiating `sin⁻¹x` and integrating `1`. The integral of `x/√(1-x²) `is then solved by substitution. If the question implies `sin⁻¹(cos x)`, we simplify `sin⁻¹(cos x)` to `π/2 - x` using a trigonometric identity, and then integrate `π/2` and `-x` separately.

Exam Tip: For inverse trigonometric functions like `sin^-1 x`, integration by parts is often required. If an argument involves `cos x`, consider using `sin^-1(cos x) = π/2 - x` for simplification before integrating.

Question 22. \( \frac{1}{\cos(x-a)\cos(x-b)} \)
Answer:
Let \( I = \int \frac{1}{\cos(x-a)\cos(x-b)} \, dx \)
We can use a trigonometric manipulation by multiplying and dividing by \( \sin(a-b) \).
\( I = \frac{1}{\sin(a-b)} \int \frac{\sin(a-b)}{\cos(x-a)\cos(x-b)} \, dx \)
Rewrite \( \sin(a-b) \) as \( \sin((x-b) - (x-a)) \):
\( I = \frac{1}{\sin(a-b)} \int \frac{\sin((x-b) - (x-a))}{\cos(x-a)\cos(x-b)} \, dx \)
Using the identity \( \sin(A-B) = \sin A \cos B - \cos A \sin B \):
\( I = \frac{1}{\sin(a-b)} \int \frac{\sin(x-b)\cos(x-a) - \cos(x-b)\sin(x-a)}{\cos(x-a)\cos(x-b)} \, dx \)
Split the fraction into two terms:
\( I = \frac{1}{\sin(a-b)} \int \left( \frac{\sin(x-b)\cos(x-a)}{\cos(x-a)\cos(x-b)} - \frac{\cos(x-b)\sin(x-a)}{\cos(x-a)\cos(x-b)} \right) dx \)
Simplify each term:
\( I = \frac{1}{\sin(a-b)} \int \left( \frac{\sin(x-b)}{\cos(x-b)} - \frac{\sin(x-a)}{\cos(x-a)} \right) dx \)
\( I = \frac{1}{\sin(a-b)} \int (\tan(x-b) - \tan(x-a)) \, dx \)
Now, integrate each term:
Recall that \( \int \tan \theta \, d\theta = \ln |\sec \theta| \) or \( -\ln |\cos \theta| \).
\( I = \frac{1}{\sin(a-b)} [-\ln |\cos(x-b)| - (-\ln |\cos(x-a)|)] + C \)
\( I = \frac{1}{\sin(a-b)} [-\ln |\cos(x-b)| + \ln |\cos(x-a)|] + C \)
Using the logarithm property \( \ln A - \ln B = \ln(A/B) \):
\( I = \frac{1}{\sin(a-b)} \ln \left| \frac{\cos(x-a)}{\cos(x-b)} \right| + C \)
In simple words: We multiply the integrand by `sin(a-b)` in both the numerator and denominator. We then rewrite the `sin(a-b)` in the numerator as `sin((x-b)-(x-a))` and use a trigonometric expansion. This allows us to split the fraction into `tan(x-b)` and `tan(x-a)`. Finally, we integrate these `tan` terms, which results in logarithmic functions.

Exam Tip: For integrals of the form `1 / (cos(x-a) cos(x-b))` or `1 / (sin(x-a) sin(x-b))`, multiply and divide by `sin(a-b)` or `cos(a-b)` respectively, and use the angle subtraction formula in the numerator to split the fraction into `tan` (or `cot`) terms.

Choose the correct answers in the following questions 23 and 24:

Question 23. \( \int \frac{\sin^2 x-\cos^2 x}{\sin^2 x \cos^2 x} \)
(A) \( \tan x + \cot x + C \)
(B) \( \tan x + \cosec x + C \)
(C) \( - \tan x + \cot x + C \)
(D) \( \tan x + \sec x + C \)
Answer: (A) \( \tan x + \cot x + C \)
Let \( I = \int \frac{\sin^2 x-\cos^2 x}{\sin^2 x \cos^2 x} \, dx \)
Split the fraction into two separate terms:
\( I = \int \left( \frac{\sin^2 x}{\sin^2 x \cos^2 x} - \frac{\cos^2 x}{\sin^2 x \cos^2 x} \right) dx \)
Simplify each term:
\( I = \int \left( \frac{1}{\cos^2 x} - \frac{1}{\sin^2 x} \right) dx \)
Rewrite in terms of `sec x` and `cosec x`:
\( I = \int (\sec^2 x - \cosec^2 x) \, dx \)
Integrate each term:
Recall that \( \int \sec^2 x \, dx = \tan x \) and \( \int \cosec^2 x \, dx = -\cot x \).
\( I = \tan x - (-\cot x) + C \)
\( I = \tan x + \cot x + C \)
Thus, option (A) is the correct choice.
In simple words: We split the given fraction into two separate fractions. Then, we simplify each fraction, turning them into `sec²x` and `cosec²x`. We integrate these standard forms; `sec²x` gives `tan x`, and `-cosec²x` gives `cot x`. Combining these results in the final answer.

Exam Tip: For integrals with a binomial numerator and a product denominator, always try splitting the fraction first. This strategy often simplifies the terms into standard integrable forms like `sec^2 x` and `cosec^2 x`.

Question 24. \( \int \frac{e^{x}(1+x)}{\cos^2(x e^{x})} \, dx \) is equal to.
(A) \( - \cot (e^x x) \)
(B) \( \tan (x e^x) + C \)
(C) \( \tan (e^x) + C \)
(D) \( \cot (e^x) + C \)
Answer: (B) \( \tan (x e^x) + C \)
Let \( I = \int \frac{e^{x}(1+x)}{\cos^2(x e^{x})} \, dx \)
This integral suggests a substitution involving the term in the argument of `cos²`.
Let \( t = x e^x \).
Now, find \( dt \) by differentiating \( t \) with respect to \( x \) using the product rule:
\( \frac{dt}{dx} = 1 \cdot e^x + x \cdot e^x = e^x(1 + x) \)
So, \( dt = e^x(1 + x) \, dx \).
Substitute \( t \) and \( dt \) into the integral:
\( I = \int \frac{dt}{\cos^2 t} \)
Rewrite \( \frac{1}{\cos^2 t} \) as \( \sec^2 t \):
\( I = \int \sec^2 t \, dt \)
Integrate \( \sec^2 t \):
\( I = \tan t + C \)
Substitute back \( t = x e^x \):
\( I = \tan (x e^x) + C \)
Thus, option (B) is the correct choice.
In simple words: We notice a complex expression `x e^x` inside the `cos²` term and its derivative `e^x(1+x)` in the numerator. This suggests using substitution. By letting `t` equal `x e^x`, the integral simplifies to `sec²t dt`, which integrates directly to `tan t`. Finally, we replace `t` with `x e^x` to get the answer.

Exam Tip: When an integral contains a function and its derivative, particularly within a more complex structure (like `cos^2(f(x))`), consider using a substitution with `t = f(x)`. This often simplifies the integral significantly.