OP Malhotra Class 12 Maths Solutions Chapter 13 Indefinite Integral 1 Exercise 13 (A)

Question 1. Integrate the following:
(i) sin 2x
(ii) 2 sin 3x
(iii) \( \frac { 1 }{ 3 } \cos 4x \)
(iv) \( \frac{\cos 5 x}{2} \)
(v) \( 8 \sec^2 8x \)
(vi) \( \operatorname{cosec}^2 2x \)
(vii) sec 5x tan 5x
(viii) -cosec 3x cot 3x
Answer:
(i) \( \int \sin 2x \,dx = \frac{-\cos 2x}{2} + C \) (Since \( \int \sin mx \,dx = \frac{-\cos mx}{m} + C \))
(ii) \( \int 2\sin 3x \,dx = \frac{-2\cos 3x}{3} + C \)
(iii) \( \int \frac{1}{3} \cos 4x \,dx = \frac{1}{3} \cdot \frac{\sin 4x}{4} + C = \frac{1}{12} \sin 4x + C \) (Since \( \int \cos mx \,dx = \frac{\sin mx}{m} + C \))
(iv) \( \int \frac{\cos 5x}{2} \,dx = \frac{1}{2} \cdot \frac{\sin 5x}{5} + C = \frac{\sin 5x}{10} + C \)
(v) \( \int 8 \sec^2 8x \,dx = 8 \cdot \frac{\tan 8x}{8} + C = \tan 8x + C \) (Since \( \int \sec^2 x \,dx = \tan x + C \))
(vi) \( \int \operatorname{cosec}^2 2x \,dx = \frac{-\cot 2x}{2} + C \) (Since \( \int \operatorname{cosec}^2 x \,dx = -\cot x + C \))
(vii) \( \int \sec 5x \tan 5x \,dx = \frac{\sec 5x}{5} + C \) (Since \( \int \sec ax \tan ax \,dx = \frac{\sec ax}{a} + C \))
(viii) \( \int -\operatorname{cosec} 3x \cot 3x \,dx = \frac{-(-\operatorname{cosec} 3x)}{3} + C = \frac{\operatorname{cosec} 3x}{3} + C \)
In simple words: To integrate basic trigonometric functions like sine, cosine, secant, and cosecant, we use their standard integration formulas. Remember to divide by the coefficient of x in the argument, and always add the constant of integration, C.

🎯 Exam Tip: When integrating functions of the form \( f(ax+b) \), remember that the integral will be \( \frac{1}{a} F(ax+b) + C \), where \( F \) is the integral of \( f \). Don't forget the constant of integration, C.

Question 2. Integrate the following:
(i) cos (5 – 3x)
(ii) \( 2\sin \left(\frac{\pi}{2}-\frac{x}{2}\right) \)
(iii) \( \sin\left(\frac{3}{4} x+5\right) \)
(iv) \( 4 \sec^2(2x - 4) \)
Answer:
(i) \( \int \cos(5-3x) \,dx = \frac{\sin(5-3x)}{-3} + C \) (Since \( \int \cos(ax+b) \,dx = \frac{\sin(ax+b)}{a} + C \))
(ii) \( \int 2\sin\left(\frac{\pi}{2}-\frac{x}{2}\right) \,dx = \int 2\cos\left(\frac{x}{2}\right) \,dx \) (Using \( \sin(\frac{\pi}{2}-\theta) = \cos \theta \))
\( \implies = 2 \frac{\sin(x/2)}{1/2} + C = 4\sin\left(\frac{x}{2}\right) + C \)
(iii) \( \int \sin\left(\frac{3}{4} x+5\right) \,dx = \frac{-\cos\left(\frac{3}{4} x+5\right)}{3/4} + C = -\frac{4}{3}\cos\left(\frac{3}{4} x+5\right) + C \)
(iv) \( \int 4\sec^2(2x-4) \,dx = 4 \frac{\tan(2x-4)}{2} + C = 2\tan(2x-4) + C \)
In simple words: For integrals involving \( ax+b \) inside the function, integrate the function as usual and then divide by 'a', the number multiplying x. Also, use basic trigonometry rules like \( \sin(\frac{\pi}{2}-\theta) = \cos \theta \) to simplify before integrating.

🎯 Exam Tip: Always look for opportunities to simplify trigonometric expressions using identities before integrating. This can make the integral much easier to solve.

Question 3. Integrate the following:
(i) \( \sin^2 x \)
(ii) \( \cos^2 x \)
(iii) \( \sin^3 x \)
(iv) \( \sin^2 mx \)
(v) \( \sin^2 x \cos^2 x \)
(vi) \( \sin^3 x \cos^3 x \)
(vii) \( \frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x} \)
(viii) \( \sin x \sec^2 x \)
(ix) \( \sin^3 x \cos^3 x \)
(x) \( \frac{1}{\sin ^2 x \cos ^2 x} \)
(xi) \( \frac{\sec x}{\sec x+\tan x} \)
(xii) \( 3 \operatorname{cosec}^2 x + 2 \sin 3x \)
Answer:
(i) \( \int \sin^2 x \,dx = \int \frac{1-\cos 2x}{2} \,dx = \frac{1}{2} \left[ \int 1\,dx - \int \cos 2x \,dx \right] = \frac{1}{2} \left[ x - \frac{\sin 2x}{2} \right] + C \)
(ii) \( \int \cos^2 x \,dx = \int \frac{1+\cos 2x}{2} \,dx = \frac{1}{2} \left[ \int 1\,dx + \int \cos 2x \,dx \right] = \frac{1}{2} \left[ x + \frac{\sin 2x}{2} \right] + C = \frac{x}{2} + \frac{\sin 2x}{4} + C \)
(iii) \( \int \sin^3 x \,dx = \int \frac{3\sin x - \sin 3x}{4} \,dx \) (Using \( \sin 3\theta = 3\sin\theta - 4\sin^3\theta \implies \sin^3\theta = \frac{1}{4}[3\sin\theta - \sin 3\theta] \))
\( \implies = \frac{1}{4} \left[ -3\cos x + \frac{\cos 3x}{3} \right] + C \)
(iv) \( \int \sin^2 mx \,dx = \int \frac{1-\cos 2mx}{2} \,dx = \frac{1}{2} \left[ x - \frac{\sin 2mx}{2m} \right] + C \) (Since \( \int \cos(ax+b) \,dx = \frac{\sin(ax+b)}{a} + C \))
(v) Let \( I = \int \sin^2 x \cos^2 x \,dx \)
\( \implies = \int \frac{1}{4} (4\sin^2 x \cos^2 x) \,dx = \frac{1}{4} \int (2\sin x \cos x)^2 \,dx \)
\( \implies = \frac{1}{4} \int \sin^2 2x \,dx = \frac{1}{4} \int \frac{1-\cos 4x}{2} \,dx \)
\( \implies = \frac{1}{8} \left[ x - \frac{\sin 4x}{4} \right] + C = \frac{x}{8} - \frac{\sin 4x}{32} + C \)
(vi) Let \( I = \int \sin^3 x \cos^3 x \,dx \)
\( \implies = \int \frac{1}{8} (8\sin^3 x \cos^3 x) \,dx = \frac{1}{8} \int (2\sin x \cos x)^3 \,dx \)
\( \implies = \frac{1}{8} \int \sin^3 2x \,dx \)
\( \implies = \frac{1}{8} \int \frac{3\sin 2x - \sin 6x}{4} \,dx \) (Using \( \sin 3\theta = 3\sin\theta - 4\sin^3\theta \))
\( \implies = \frac{1}{32} \left[ -3\frac{\cos 2x}{2} + \frac{\cos 6x}{6} \right] + C = -\frac{3\cos 2x}{64} + \frac{\cos 6x}{192} + C \)
(vii) Let \( I = \int \frac{\cos 2x+2 \sin ^2 x}{\cos ^2 x} \,dx \)
\( \implies = \int \frac{1-2\sin^2 x + 2\sin^2 x}{\cos ^2 x} \,dx \) (Using \( \cos 2x = 1-2\sin^2 x \))
\( \implies = \int \frac{1}{\cos ^2 x} \,dx = \int \sec^2 x \,dx = \tan x + C \)
(viii) \( \int \sin x \sec^2 x \,dx = \int \sin x \cdot \frac{1}{\cos x} \cdot \frac{1}{\cos x} \,dx = \int \tan x \sec x \,dx = \sec x + C \)
(ix) This is the same as (vi). The solution is: \( -\frac{3\cos 2x}{64} + \frac{\cos 6x}{192} + C \)
(x) Let \( I = \int \frac{1}{\sin ^2 x \cos ^2 x} \,dx \)
\( \implies = \int \frac{\sin^2 x + \cos^2 x}{\sin ^2 x \cos ^2 x} \,dx \) (Using \( \sin^2 x + \cos^2 x = 1 \))
\( \implies = \int \left( \frac{\sin^2 x}{\sin^2 x \cos^2 x} + \frac{\cos^2 x}{\sin^2 x \cos^2 x} \right) \,dx = \int \left( \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} \right) \,dx \)
\( \implies = \int (\sec^2 x + \operatorname{cosec}^2 x) \,dx = \tan x - \cot x + C \)
(xi) Let \( I = \int \frac{\sec x}{\sec x+\tan x} \,dx \)
\( \implies = \int \frac{\sec x (\sec x - \tan x)}{(\sec x+\tan x)(\sec x - \tan x)} \,dx = \int \frac{\sec^2 x - \sec x \tan x}{\sec^2 x - \tan^2 x} \,dx \)
\( \implies = \int \frac{\sec^2 x - \sec x \tan x}{1} \,dx \) (Using \( \sec^2 x - \tan^2 x = 1 \))
\( \implies = \int (\sec^2 x - \sec x \tan x) \,dx = \tan x - \sec x + C \)
(xii) \( \int (3 \operatorname{cosec}^2 x + 2 \sin 3x) \,dx = 3 \int \operatorname{cosec}^2 x \,dx + 2 \int \sin 3x \,dx \)
\( \implies = 3(-\cot x) + 2\left(-\frac{\cos 3x}{3}\right) + C = -3\cot x - \frac{2}{3}\cos 3x + C \)
In simple words: For higher powers of sine and cosine, use power-reducing formulas like \( \sin^2 x = \frac{1-\cos 2x}{2} \) or trigonometric identities like \( \sin 3x \) to change them into simpler forms that are easy to integrate. Also, sometimes simplifying the expression by converting to sine and cosine or using algebraic manipulation helps a lot. Remember that \( \sin^2 x + \cos^2 x = 1 \) is a very useful identity for simplifying fractions.

🎯 Exam Tip: Mastering trigonometric identities is crucial for solving integration problems involving powers of trigonometric functions. Always keep in mind the identities for \( \sin^2 x \), \( \cos^2 x \), \( \sin^3 x \), \( \cos^3 x \), \( \sec^2 x - \tan^2 x = 1 \), and \( \operatorname{cosec}^2 x - \cot^2 x = 1 \).

Question 4. Integrate the following:
(i) cos 4x cos 3x dx
(ii) sin 4x sin 8x dx
Answer:
(i) Let \( I = \int \cos 4x \cos 3x \,dx \)
\( \implies = \frac{1}{2} \int (2\cos 4x \cos 3x) \,dx \)
\( \implies = \frac{1}{2} \int [\cos(4x+3x) + \cos(4x-3x)] \,dx \) (Using \( 2\cos A \cos B = \cos(A+B) + \cos(A-B) \))
\( \implies = \frac{1}{2} \int [\cos 7x + \cos x] \,dx \)
\( \implies = \frac{1}{2} \left[ \frac{\sin 7x}{7} + \sin x \right] + C \)
(ii) Let \( I = \int \sin 4x \sin 8x \,dx \)
\( \implies = \frac{1}{2} \int (2\sin 8x \sin 4x) \,dx \)
\( \implies = \frac{1}{2} \int [\cos(8x-4x) - \cos(8x+4x)] \,dx \) (Using \( 2\sin A \sin B = \cos(A-B) - \cos(A+B) \))
\( \implies = \frac{1}{2} \int [\cos 4x - \cos 12x] \,dx \)
\( \implies = \frac{1}{2} \left[ \frac{\sin 4x}{4} - \frac{\sin 12x}{12} \right] + C \)
In simple words: When you need to integrate products of sine and cosine functions, convert them into sums or differences using trigonometric product-to-sum formulas. This makes them much easier to integrate because sums are integrated term by term.

🎯 Exam Tip: Remember the product-to-sum trigonometric identities: \( 2\cos A \cos B \), \( 2\sin A \sin B \), \( 2\sin A \cos B \), and \( 2\cos A \sin B \). These are essential for converting products into integrable forms.

Question 5. Integrate the following:
(i) \( \cos 2x \cos 4x \cos 6x \)
(ii) \( \sin x \sin 2x \sin 3x \)
(iii) \( \frac{\cos ^2 x-\sin ^2 x}{\sqrt{1+\cos 4 x}} \)
(iv) \( \cos^4 x \sin^4 x \)
(v) \( \frac{7 \cos ^3 x+4 \sin ^3 x}{3 \sin ^2 x \cos ^2 x} \)
Answer:
(i) Let \( I = \int \cos 2x \cos 4x \cos 6x \,dx \)
\( \implies = \frac{1}{2} \int (2\cos 4x \cos 2x) \cos 6x \,dx \)
\( \implies = \frac{1}{2} \int [\cos(4x+2x) + \cos(4x-2x)] \cos 6x \,dx \)
\( \implies = \frac{1}{2} \int (\cos 6x + \cos 2x) \cos 6x \,dx \)
\( \implies = \frac{1}{2} \int (\cos^2 6x + \cos 6x \cos 2x) \,dx \)
\( \implies = \frac{1}{2} \int \frac{1+\cos 12x}{2} \,dx + \frac{1}{2} \cdot \frac{1}{2} \int (2\cos 6x \cos 2x) \,dx \)
\( \implies = \frac{1}{4} \left[ x + \frac{\sin 12x}{12} \right] + \frac{1}{4} \int [\cos(6x+2x) + \cos(6x-2x)] \,dx \)
\( \implies = \frac{1}{4} \left[ x + \frac{\sin 12x}{12} \right] + \frac{1}{4} \int [\cos 8x + \cos 4x] \,dx \)
\( \implies = \frac{1}{4} \left[ x + \frac{\sin 12x}{12} \right] + \frac{1}{4} \left[ \frac{\sin 8x}{8} + \frac{\sin 4x}{4} \right] + C \)
(ii) Let \( I = \int \sin x \sin 2x \sin 3x \,dx \)
\( \implies = \frac{1}{2} \int (2\sin 2x \sin x) \sin 3x \,dx \)
\( \implies = \frac{1}{2} \int [\cos(2x-x) - \cos(2x+x)] \sin 3x \,dx \)
\( \implies = \frac{1}{2} \int (\cos x - \cos 3x) \sin 3x \,dx \)
\( \implies = \frac{1}{2} \int (\cos x \sin 3x - \cos 3x \sin 3x) \,dx \)
\( \implies = \frac{1}{2} \int \left( \frac{1}{2}(2\sin 3x \cos x) - \frac{1}{2}(2\sin 3x \cos 3x) \right) \,dx \)
\( \implies = \frac{1}{4} \int [\sin(3x+x) + \sin(3x-x)] \,dx - \frac{1}{4} \int \sin 6x \,dx \)
\( \implies = \frac{1}{4} \int [\sin 4x + \sin 2x] \,dx - \frac{1}{4} \int \sin 6x \,dx \)
\( \implies = \frac{1}{4} \left[ -\frac{\cos 4x}{4} - \frac{\cos 2x}{2} \right] - \frac{1}{4} \left[ -\frac{\cos 6x}{6} \right] + C \)
(iii) Let \( I = \int \frac{\cos ^2 x-\sin ^2 x}{\sqrt{1+\cos 4 x}} \,dx \)
\( \implies = \int \frac{\cos 2x}{\sqrt{2\cos^2 2x}} \,dx \) (Using \( \cos^2 x - \sin^2 x = \cos 2x \) and \( 1+\cos 4x = 2\cos^2 2x \))
\( \implies = \int \frac{\cos 2x}{\sqrt{2} |\cos 2x|} \,dx \)
Assuming \( \cos 2x > 0 \), \( \implies = \int \frac{1}{\sqrt{2}} \,dx = \frac{1}{\sqrt{2}} x + C \)
(iv) Let \( I = \int \cos^4 x \sin^4 x \,dx \)
\( \implies = \int \frac{1}{16} (16\sin^4 x \cos^4 x) \,dx = \frac{1}{16} \int (2\sin x \cos x)^4 \,dx \)
\( \implies = \frac{1}{16} \int \sin^4 2x \,dx \)
\( \implies = \frac{1}{16} \int \left( \frac{1-\cos 4x}{2} \right)^2 \,dx = \frac{1}{16} \int \frac{1+\cos^2 4x - 2\cos 4x}{4} \,dx \)
\( \implies = \frac{1}{64} \int \left( 1 + \frac{1+\cos 8x}{2} - 2\cos 4x \right) \,dx \)
\( \implies = \frac{1}{64} \int \left( \frac{3}{2} + \frac{\cos 8x}{2} - 2\cos 4x \right) \,dx \)
\( \implies = \frac{1}{128} \int (3 + \cos 8x - 4\cos 4x) \,dx \)
\( \implies = \frac{1}{128} \left[ 3x + \frac{\sin 8x}{8} - 4\frac{\sin 4x}{4} \right] + C = \frac{1}{128} \left[ 3x + \frac{\sin 8x}{8} - \sin 4x \right] + C \)
(v) Let \( I = \int \frac{7 \cos ^3 x+4 \sin ^3 x}{3 \sin ^2 x \cos ^2 x} \,dx \)
\( \implies = \int \left( \frac{7 \cos^3 x}{3 \sin^2 x \cos^2 x} + \frac{4 \sin^3 x}{3 \sin^2 x \cos^2 x} \right) \,dx \)
\( \implies = \int \left( \frac{7 \cos x}{3 \sin^2 x} + \frac{4 \sin x}{3 \cos^2 x} \right) \,dx \)
\( \implies = \frac{7}{3} \int \cot x \operatorname{cosec} x \,dx + \frac{4}{3} \int \tan x \sec x \,dx \)
\( \implies = \frac{7}{3} (-\operatorname{cosec} x) + \frac{4}{3} (\sec x) + C \)
In simple words: For complex integrals, break them down using trigonometric identities. For products of many terms, use product-to-sum formulas repeatedly. For expressions involving sums in the numerator and products in the denominator, split the fraction into simpler terms. High powers of sine and cosine can be simplified using double-angle or power-reducing formulas. Remember to handle square roots carefully with absolute values.

🎯 Exam Tip: When dealing with products of three or more trigonometric functions, apply the product-to-sum identities stepwise. For fractions, try to split the numerator or convert all terms to sine and cosine to simplify. Double-check your algebraic manipulations and trigonometric identities.

Question 6. Integrate the following:
(i) \( \frac{1}{1+\cos x} \)
(ii) \( \frac{1}{1-\cos 2 x} \)
(iii) \( \frac{1}{1-\sin x} \)
(iv) \( \frac{1-\cos 2 x}{1+\cos 2 x} \)
(v) \( \sqrt{1+\cos x} \)
(vi) \( \sqrt{1+\sin 2 x} \)
(vii) \( \cos x \sqrt{1+\cos 2 x} \)
(viii) \( \sin x \sqrt{1-\cos 2 x} \)
(ix) \( \frac{\cos x-\sin x}{\cos x+\sin x}(2+2 \sin 2 x) \)
(x) \( \frac{4-5 \sin x}{\cos ^2 x}+\frac{1}{\sin ^2 x \cos ^2 x} \)
(xi) \( \frac{\sin x+\cos x}{\sqrt{1+\sin 2 x}} \)
Answer:
(i) Let \( I = \int \frac{1}{1+\cos x} \,dx \)
\( \implies = \int \frac{1}{2\cos^2 (x/2)} \,dx \) (Using \( 1+\cos x = 2\cos^2 (x/2) \))
\( \implies = \frac{1}{2} \int \sec^2 (x/2) \,dx = \frac{1}{2} \frac{\tan(x/2)}{1/2} + C = \tan(x/2) + C \)
(ii) Let \( I = \int \frac{1}{1-\cos 2x} \,dx \)
\( \implies = \int \frac{1}{2\sin^2 x} \,dx \) (Using \( 1-\cos 2x = 2\sin^2 x \))
\( \implies = \frac{1}{2} \int \operatorname{cosec}^2 x \,dx = \frac{1}{2} (-\cot x) + C = -\frac{\cot x}{2} + C \)
(iii) Let \( I = \int \frac{1}{1-\sin x} \,dx \)
\( \implies = \int \frac{1}{1-\sin x} \cdot \frac{1+\sin x}{1+\sin x} \,dx = \int \frac{1+\sin x}{1-\sin^2 x} \,dx = \int \frac{1+\sin x}{\cos^2 x} \,dx \)
\( \implies = \int \left( \frac{1}{\cos^2 x} + \frac{\sin x}{\cos^2 x} \right) \,dx = \int (\sec^2 x + \tan x \sec x) \,dx \)
\( \implies = \tan x + \sec x + C \)
(iv) Let \( I = \int \frac{1-\cos 2 x}{1+\cos 2 x} \,dx \)
\( \implies = \int \frac{2\sin^2 x}{2\cos^2 x} \,dx \) (Using \( 1-\cos 2x = 2\sin^2 x \) and \( 1+\cos 2x = 2\cos^2 x \))
\( \implies = \int \tan^2 x \,dx = \int (\sec^2 x - 1) \,dx \)
\( \implies = \tan x - x + C \)
(v) Let \( I = \int \sqrt{1+\cos x} \,dx \)
\( \implies = \int \sqrt{2\cos^2 (x/2)} \,dx \) (Using \( 1+\cos x = 2\cos^2 (x/2) \))
\( \implies = \int \sqrt{2} |\cos(x/2)| \,dx \)
Assuming \( \cos(x/2) > 0 \), \( \implies = \sqrt{2} \int \cos(x/2) \,dx = \sqrt{2} \frac{\sin(x/2)}{1/2} + C = 2\sqrt{2}\sin(x/2) + C \)
(vi) Let \( I = \int \sqrt{1+\sin 2x} \,dx \)
\( \implies = \int \sqrt{\sin^2 x + \cos^2 x + 2\sin x \cos x} \,dx \) (Using \( 1 = \sin^2 x + \cos^2 x \))
\( \implies = \int \sqrt{(\sin x + \cos x)^2} \,dx = \int |\sin x + \cos x| \,dx \)
Assuming \( \sin x + \cos x > 0 \), \( \implies = \int (\sin x + \cos x) \,dx = -\cos x + \sin x + C \)
(vii) Let \( I = \int \cos x \sqrt{1+\cos 2x} \,dx \)
\( \implies = \int \cos x \sqrt{2\cos^2 x} \,dx \) (Using \( 1+\cos 2x = 2\cos^2 x \))
\( \implies = \int \cos x \sqrt{2} |\cos x| \,dx \)
Assuming \( \cos x > 0 \), \( \implies = \sqrt{2} \int \cos^2 x \,dx = \sqrt{2} \int \frac{1+\cos 2x}{2} \,dx \)
\( \implies = \frac{\sqrt{2}}{2} \left[ x + \frac{\sin 2x}{2} \right] + C \)
(viii) Let \( I = \int \sin x \sqrt{1-\cos 2x} \,dx \)
\( \implies = \int \sin x \sqrt{2\sin^2 x} \,dx \) (Using \( 1-\cos 2x = 2\sin^2 x \))
\( \implies = \int \sin x \sqrt{2} |\sin x| \,dx \)
Assuming \( \sin x > 0 \), \( \implies = \sqrt{2} \int \sin^2 x \,dx = \sqrt{2} \int \frac{1-\cos 2x}{2} \,dx \)
\( \implies = \frac{\sqrt{2}}{2} \left[ x - \frac{\sin 2x}{2} \right] + C \)
(ix) Let \( I = \int \frac{\cos x-\sin x}{\cos x+\sin x}(2+2 \sin 2 x) \,dx \)
\( \implies = \int \frac{\cos x-\sin x}{\cos x+\sin x} \cdot 2(\sin^2 x + \cos^2 x + 2\sin x \cos x) \,dx \)
\( \implies = \int \frac{\cos x-\sin x}{\cos x+\sin x} \cdot 2(\cos x + \sin x)^2 \,dx \)
\( \implies = \int 2(\cos x-\sin x)(\cos x + \sin x) \,dx \)
\( \implies = \int 2(\cos^2 x - \sin^2 x) \,dx = \int 2\cos 2x \,dx \)
\( \implies = 2\frac{\sin 2x}{2} + C = \sin 2x + C \)
(x) Let \( I = \int \left( \frac{4-5 \sin x}{\cos ^2 x}+\frac{1}{\sin ^2 x \cos ^2 x} \right) \,dx \)
\( \implies = \int \left( \frac{4}{\cos^2 x} - \frac{5 \sin x}{\cos^2 x} + \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} \right) \,dx \)
\( \implies = \int \left( 4\sec^2 x - 5\tan x \sec x + \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} \right) \,dx \)
\( \implies = \int (4\sec^2 x - 5\tan x \sec x + \sec^2 x + \operatorname{cosec}^2 x) \,dx \)
\( \implies = \int (5\sec^2 x - 5\tan x \sec x + \operatorname{cosec}^2 x) \,dx \)
\( \implies = 5\tan x - 5\sec x - \cot x + C \)
(xi) Let \( I = \int \frac{\sin x+\cos x}{\sqrt{1+\sin 2 x}} \,dx \)
\( \implies = \int \frac{\sin x+\cos x}{\sqrt{(\sin x+\cos x)^2}} \,dx \) (Using \( 1+\sin 2x = (\sin x+\cos x)^2 \))
\( \implies = \int \frac{\sin x+\cos x}{|\sin x+\cos x|} \,dx \)
Assuming \( \sin x+\cos x > 0 \), \( \implies = \int 1 \,dx = x + C \)
In simple words: Many integrals can be simplified using basic trigonometric identities before integration. For fractions, try to use identities like \( 1+\cos x = 2\cos^2 (x/2) \) or rationalize the denominator. Square roots often involve identities like \( 1+\sin 2x = (\sin x+\cos x)^2 \). When simplifying, be mindful of absolute values, as square roots always result in non-negative values.

🎯 Exam Tip: Always look for ways to simplify the integrand using half-angle, double-angle, or basic identities before attempting integration. This often transforms a complicated integral into a standard one. Pay close attention to conditions like \( \cos x > 0 \) or \( \sin x+\cos x > 0 \) if you remove absolute value signs from square roots.

Question 7. Integrate the following:
\( \sqrt{\left(1+\sin \frac{x}{2}\right)} \)
Answer:
Let \( I = \int \sqrt{1+\sin \frac{x}{2}} \,dx \)
\( \implies = \int \sqrt{\sin^2 \frac{x}{4} + \cos^2 \frac{x}{4} + 2\sin \frac{x}{4} \cos \frac{x}{4}} \,dx \) (Using \( 1 = \sin^2 \theta + \cos^2 \theta \) and \( \sin 2\theta = 2\sin\theta \cos\theta \))
\( \implies = \int \sqrt{\left(\sin \frac{x}{4} + \cos \frac{x}{4}\right)^2} \,dx = \int \left|\sin \frac{x}{4} + \cos \frac{x}{4}\right| \,dx \)
Assuming \( \sin \frac{x}{4} + \cos \frac{x}{4} > 0 \), \( \implies = \int \left(\sin \frac{x}{4} + \cos \frac{x}{4}\right) \,dx \)
\( \implies = \frac{-\cos(x/4)}{1/4} + \frac{\sin(x/4)}{1/4} + C = -4\cos\left(\frac{x}{4}\right) + 4\sin\left(\frac{x}{4}\right) + C \)
In simple words: To integrate a square root like this, use the identity \( 1 = \sin^2 \theta + \cos^2 \theta \) and \( \sin 2\theta = 2\sin\theta \cos\theta \) to turn the expression inside the root into a perfect square. This allows you to remove the square root and integrate the simpler sum of sine and cosine functions. Remember to adjust the angle properly.

🎯 Exam Tip: When \( 1+\sin A \) appears inside a square root, always aim to convert it into \( (\sin(A/2) + \cos(A/2))^2 \) using \( 1 = \sin^2(A/2) + \cos^2(A/2) \) and \( \sin A = 2\sin(A/2)\cos(A/2) \).

Question 8. Integrate the following:
\( \frac{\sin ^6 x+\cos ^6 x}{\sin ^2 x \cos ^2 x} \)
Answer:
Let \( I = \int \frac{\sin ^6 x+\cos ^6 x}{\sin ^2 x \cos ^2 x} \,dx \)
We know \( a^3+b^3 = (a+b)(a^2-ab+b^2) \). Let \( a = \sin^2 x \) and \( b = \cos^2 x \).
\( \implies \sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) \)
\( \implies = (1)((\sin^2 x + \cos^2 x)^2 - 3\sin^2 x \cos^2 x) \)
\( \implies = 1^2 - 3\sin^2 x \cos^2 x = 1 - 3\sin^2 x \cos^2 x \)
Now, substitute this back into the integral:
\( \implies I = \int \frac{1 - 3\sin^2 x \cos^2 x}{\sin ^2 x \cos ^2 x} \,dx \)
\( \implies = \int \left( \frac{1}{\sin^2 x \cos^2 x} - \frac{3\sin^2 x \cos^2 x}{\sin^2 x \cos^2 x} \right) \,dx \)
\( \implies = \int \left( \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} - 3 \right) \,dx \) (Using \( 1 = \sin^2 x + \cos^2 x \))
\( \implies = \int \left( \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} - 3 \right) \,dx \)
\( \implies = \int (\sec^2 x + \operatorname{cosec}^2 x - 3) \,dx \)
\( \implies = \tan x - \cot x - 3x + C \)
In simple words: For higher even powers like \( \sin^6 x + \cos^6 x \), use algebraic identities like \( a^3+b^3 \) and trigonometric identity \( \sin^2 x + \cos^2 x = 1 \) to simplify the numerator. Then, split the fraction into simpler terms that can be directly integrated. This method breaks down complex expressions into manageable parts.

🎯 Exam Tip: Always try to simplify complex rational trigonometric functions using identities like \( a^3+b^3 \) or \( (a+b)^2 \). Look for opportunities to introduce \( \sin^2 x + \cos^2 x = 1 \) to reduce the expression to secant and cosecant terms.

Question 9. Integrate the following:
\( \sin^6 x \)
Answer:
Let \( I = \int \sin^6 x \,dx = \int (\sin^3 x)^2 \,dx \)
We know \( \sin 3x = 3\sin x - 4\sin^3 x \implies \sin^3 x = \frac{3\sin x - \sin 3x}{4} \)
\( \implies I = \int \left[ \frac{3\sin x - \sin 3x}{4} \right]^2 \,dx = \frac{1}{16} \int (9\sin^2 x - 6\sin x \sin 3x + \sin^2 3x) \,dx \)
Now, we integrate each term:
\( \int 9\sin^2 x \,dx = 9 \int \frac{1-\cos 2x}{2} \,dx = \frac{9}{2} \left( x - \frac{\sin 2x}{2} \right) \)
\( \int \sin^2 3x \,dx = \int \frac{1-\cos 6x}{2} \,dx = \frac{1}{2} \left( x - \frac{\sin 6x}{6} \right) \)
\( \int 6\sin x \sin 3x \,dx = 3 \int (2\sin 3x \sin x) \,dx \)
\( \implies = 3 \int [\cos(3x-x) - \cos(3x+x)] \,dx = 3 \int (\cos 2x - \cos 4x) \,dx \)
\( \implies = 3 \left( \frac{\sin 2x}{2} - \frac{\sin 4x}{4} \right) \)
Combine these terms:
\( \implies I = \frac{1}{16} \left[ \frac{9x}{2} - \frac{9\sin 2x}{4} - 3\left(\frac{\sin 2x}{2} - \frac{\sin 4x}{4}\right) + \frac{x}{2} - \frac{\sin 6x}{12} \right] + C \)
\( \implies = \frac{1}{16} \left[ \frac{9x}{2} - \frac{9\sin 2x}{4} - \frac{3\sin 2x}{2} + \frac{3\sin 4x}{4} + \frac{x}{2} - \frac{\sin 6x}{12} \right] + C \)
\( \implies = \frac{1}{16} \left[ \left(\frac{9}{2}+\frac{1}{2}\right)x - \left(\frac{9}{4}+\frac{3}{2}\right)\sin 2x + \frac{3}{4}\sin 4x - \frac{1}{12}\sin 6x \right] + C \)
\( \implies = \frac{1}{16} \left[ 5x - \frac{15}{4}\sin 2x + \frac{3}{4}\sin 4x - \frac{1}{12}\sin 6x \right] + C \)
\( \implies = \frac{1}{32} \left[ 10x - \frac{15}{2}\sin 2x + \frac{3}{2}\sin 4x - \frac{1}{6}\sin 6x \right] + C \)
In simple words: To integrate high powers of sine or cosine, use the identity for \( \sin^3 x \) (or \( \cos^3 x \)) and then square the expression. This will lead to terms that can be integrated using power-reducing formulas for \( \sin^2 x \) (or \( \cos^2 x \)) and product-to-sum formulas for \( \sin x \sin 3x \). Combine the results to get the final answer.

🎯 Exam Tip: When integrating \( \sin^n x \) or \( \cos^n x \) for even \( n \), reduce the power using \( \sin^2 x = (1-\cos 2x)/2 \) or \( \cos^2 x = (1+\cos 2x)/2 \). For odd powers, use \( \sin^2 x = 1-\cos^2 x \) and substitution, or use triple angle formulas as shown here for \( \sin^3 x \).

Question 10. Integrate the following:
\( \tan^{-1}\left(\frac{\sin 2 x}{1+\cos 2 x}\right) \)
Answer:
Let \( I = \int \tan^{-1}\left(\frac{\sin 2 x}{1+\cos 2 x}\right) \,dx \)
We know \( \sin 2x = 2\sin x \cos x \) and \( 1+\cos 2x = 2\cos^2 x \).
\( \implies I = \int \tan^{-1}\left(\frac{2\sin x \cos x}{2\cos^2 x}\right) \,dx \)
\( \implies = \int \tan^{-1}\left(\frac{\sin x}{\cos x}\right) \,dx = \int \tan^{-1}(\tan x) \,dx \)
\( \implies = \int x \,dx \)
\( \implies = \frac{x^2}{2} + C \)
In simple words: When you see an inverse tangent with a fraction inside, try to simplify the fraction using trigonometric identities. Often, the fraction will simplify to \( \tan x \), which allows you to cancel out the \( \tan^{-1} \) and integrate a much simpler expression like x.

🎯 Exam Tip: For inverse trigonometric integrals, always try to simplify the argument of the inverse function first. Use standard trigonometric identities like double-angle formulas to reduce the expression inside to a simple trigonometric ratio that matches the inverse function (e.g., \( \tan^{-1}(\tan x) = x \)).

Question 11. Integrate the following:
\( \cos^{-1}\left(\frac{1-\tan ^2 x}{1+\tan ^2 x}\right) \)
Answer:
Let \( I = \int \cos^{-1}\left(\frac{1-\tan ^2 x}{1+\tan ^2 x}\right) \,dx \)
We know that \( \cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x} \).
\( \implies I = \int \cos^{-1}(\cos 2x) \,dx \)
\( \implies = \int 2x \,dx \)
\( \implies = \frac{2x^2}{2} + C = x^2 + C \)
In simple words: When faced with an inverse cosine integral, simplify the expression inside the inverse cosine. Recognize that \( \frac{1-\tan^2 x}{1+\tan^2 x} \) is a known identity for \( \cos 2x \). Once simplified, the inverse function cancels, leaving a straightforward polynomial to integrate.

🎯 Exam Tip: Another key identity for inverse trigonometric integrals is \( \cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x} \). Knowing this and similar identities (e.g., for \( \sin 2x \) and \( \tan 2x \)) in terms of \( \tan x \) can quickly simplify complex-looking integrals.

Question 12. Integrate the following:
\( \cos^{-1}(\sin x) \)
Answer:
Let \( I = \int \cos^{-1}(\sin x) \,dx \)
We know \( \sin x = \cos\left(\frac{\pi}{2}-x\right) \).
\( \implies I = \int \cos^{-1}\left(\cos\left(\frac{\pi}{2}-x\right)\right) \,dx \)
\( \implies = \int \left(\frac{\pi}{2}-x\right) \,dx \)
\( \implies = \frac{\pi}{2} x - \frac{x^2}{2} + C \)
In simple words: To integrate an inverse cosine of a sine function, convert the sine function into a cosine function using the identity \( \sin x = \cos(\frac{\pi}{2}-x) \). This lets the inverse cosine cancel out, leaving a simple polynomial to integrate.

🎯 Exam Tip: If you have an inverse trigonometric function whose argument is a different trigonometric function (e.g., \( \cos^{-1}(\sin x) \)), try to convert the argument to match the inverse function using complementary angle identities (like \( \sin x = \cos(\frac{\pi}{2}-x) \)) or other basic transformations.

Question 7. Integrate the following: \( \sqrt{\left(1+\sin \frac{x}{2}\right)} \)
Answer: Let the integral be \( I \).
\( I = \int \sqrt{1+\sin \frac{x}{2}} dx \)
We know that \( 1 = \sin^2 \frac{x}{4} + \cos^2 \frac{x}{4} \) and \( \sin \frac{x}{2} = 2 \sin \frac{x}{4} \cos \frac{x}{4} \).
So, \( I = \int \sqrt{\sin^2 \frac{x}{4} + \cos^2 \frac{x}{4} + 2 \sin \frac{x}{4} \cos \frac{x}{4}} dx \)
\( I = \int \sqrt{\left(\sin \frac{x}{4} + \cos \frac{x}{4}\right)^2} dx \)
\( I = \int \left(\sin \frac{x}{4} + \cos \frac{x}{4}\right) dx \)
Now, we integrate term by term.
\( I = \frac{-\cos \frac{x}{4}}{1/4} + \frac{\sin \frac{x}{4}}{1/4} + C \)
\( I = -4 \cos \frac{x}{4} + 4 \sin \frac{x}{4} + C \)
\( I = 4 \left(\sin \frac{x}{4} - \cos \frac{x}{4}\right) + C \)
In simple words: First, rewrite the expression inside the square root using trigonometric identities. This helps simplify the whole term into a perfect square. Once simplified, integrate the sine and cosine parts separately.

🎯 Exam Tip: When integrating expressions with square roots involving 1 and sine or cosine, always look for half-angle formulas or perfect square identities to simplify the term under the root.

Question 8. Integrate the following: \( \frac{\sin ^6 x+\cos ^6 x}{\sin ^2 x \cos ^2 x} \)
Answer: Let the integral be \( I \).
\( I = \int \frac{\sin^6 x + \cos^6 x}{\sin^2 x \cos^2 x} dx \)
We use the identity \( a^3 + b^3 = (a+b)(a^2-ab+b^2) \). Here, let \( a = \sin^2 x \) and \( b = \cos^2 x \).
So, \( \sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) \)
We know \( \sin^2 x + \cos^2 x = 1 \).
Also, \( \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x = 1^2 - 2 \sin^2 x \cos^2 x = 1 - 2 \sin^2 x \cos^2 x \).
Substituting these into the numerator:
\( \sin^6 x + \cos^6 x = 1 \cdot ( (1 - 2 \sin^2 x \cos^2 x) - \sin^2 x \cos^2 x ) \)
\( = 1 - 3 \sin^2 x \cos^2 x \)
Now, substitute this back into the integral:
\( I = \int \frac{1 - 3 \sin^2 x \cos^2 x}{\sin^2 x \cos^2 x} dx \)
We can split the fraction:
\( I = \int \left(\frac{1}{\sin^2 x \cos^2 x} - \frac{3 \sin^2 x \cos^2 x}{\sin^2 x \cos^2 x}\right) dx \)
\( I = \int \left(\frac{1}{\sin^2 x \cos^2 x} - 3\right) dx \)
We know \( 1 = \sin^2 x + \cos^2 x \). Substitute this into the first term:
\( I = \int \left(\frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} - 3\right) dx \)
Split this fraction again:
\( I = \int \left(\frac{\sin^2 x}{\sin^2 x \cos^2 x} + \frac{\cos^2 x}{\sin^2 x \cos^2 x} - 3\right) dx \)
\( I = \int \left(\frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} - 3\right) dx \)
\( I = \int (\sec^2 x + \operatorname{cosec}^2 x - 3) dx \)
Now, integrate each term:
\( I = \tan x - \cot x - 3x + C \)
In simple words: We rewrite the top part of the fraction using a special math rule that simplifies it a lot. Then, we split the fraction into smaller, easier pieces. After that, we replace some terms with other forms like 'secant squared' and 'cosecant squared'. Finally, we integrate each simple piece to get the answer.

🎯 Exam Tip: Memorizing trigonometric identities like \( \sin^6 x + \cos^6 x = 1 - 3 \sin^2 x \cos^2 x \) and \( \sec^2 x + \operatorname{cosec}^2 x = \frac{1}{\sin^2 x \cos^2 x} \) is crucial for quickly solving such integration problems.

Question 9. Integrate the following: \( \sin^6 x \)
Answer: Let the integral be \( I \).
\( I = \int \sin^6 x dx \)
We can write \( \sin^6 x = (\sin^3 x)^2 \).
We use the triple angle identity for sine: \( \sin 3x = 3 \sin x - 4 \sin^3 x \).
From this, we get \( 4 \sin^3 x = 3 \sin x - \sin 3x \).
So, \( \sin^3 x = \frac{3 \sin x - \sin 3x}{4} \).
Substitute this into the integral:
\( I = \int \left(\frac{3 \sin x - \sin 3x}{4}\right)^2 dx \)
\( I = \frac{1}{16} \int (3 \sin x - \sin 3x)^2 dx \)
\( I = \frac{1}{16} \int (9 \sin^2 x - 6 \sin x \sin 3x + \sin^2 3x) dx \)
Now, we use power-reducing formulas: \( \sin^2 A = \frac{1 - \cos 2A}{2} \).
Also, use the product-to-sum formula: \( 2 \sin A \sin B = \cos(A-B) - \cos(A+B) \).
For \( 9 \sin^2 x \), we have \( 9 \left(\frac{1 - \cos 2x}{2}\right) = \frac{9}{2} - \frac{9}{2} \cos 2x \).
For \( 6 \sin x \sin 3x \), we have \( 3 (2 \sin x \sin 3x) = 3 (\cos(x-3x) - \cos(x+3x)) = 3 (\cos(-2x) - \cos(4x)) = 3 (\cos 2x - \cos 4x) \).
For \( \sin^2 3x \), we have \( \frac{1 - \cos(2 \cdot 3x)}{2} = \frac{1 - \cos 6x}{2} = \frac{1}{2} - \frac{1}{2} \cos 6x \).
Substitute these back into the integral:
\( I = \frac{1}{16} \int \left( \left(\frac{9}{2} - \frac{9}{2} \cos 2x\right) - \left(3 \cos 2x - 3 \cos 4x\right) + \left(\frac{1}{2} - \frac{1}{2} \cos 6x\right) \right) dx \)
\( I = \frac{1}{16} \int \left( \frac{9}{2} - \frac{9}{2} \cos 2x - 3 \cos 2x + 3 \cos 4x + \frac{1}{2} - \frac{1}{2} \cos 6x \right) dx \)
Combine the constant terms and terms with the same cosine angles:
\( I = \frac{1}{16} \int \left( \left(\frac{9}{2} + \frac{1}{2}\right) - \left(\frac{9}{2} + 3\right) \cos 2x + 3 \cos 4x - \frac{1}{2} \cos 6x \right) dx \)
\( I = \frac{1}{16} \int \left( 5 - \frac{15}{2} \cos 2x + 3 \cos 4x - \frac{1}{2} \cos 6x \right) dx \)
To remove the fraction \( \frac{1}{2} \) from inside, we can multiply and divide by 2:
\( I = \frac{1}{32} \int \left( 10 - 15 \cos 2x + 6 \cos 4x - \cos 6x \right) dx \)
Now, integrate each term:
\( I = \frac{1}{32} \left[ 10x - 15 \frac{\sin 2x}{2} + 6 \frac{\sin 4x}{4} - \frac{\sin 6x}{6} \right] + C \)
\( I = \frac{1}{32} \left[ 10x - \frac{15}{2} \sin 2x + \frac{3}{2} \sin 4x - \frac{1}{6} \sin 6x \right] + C \)
In simple words: To integrate sine to the power of six, we first change it using a rule about sine to the power of three. Then we expand this squared term. Next, we use other special math rules to make the squared sine terms into simple cosine terms and the product of sines into a difference of cosines. Finally, we combine everything and integrate each simple term.

🎯 Exam Tip: Integrating higher powers of sine or cosine often requires using multiple trigonometric identities (power-reducing, triple angle, product-to-sum) in sequence. Keep track of constant factors and arguments of the functions.

Question 10. Integrate the following: \( \tan^{-1}\left(\frac{\sin 2 x}{1+\cos 2 x}\right) \)
Answer: Let the integral be \( I \).
\( I = \int \tan^{-1}\left(\frac{\sin 2 x}{1+\cos 2 x}\right) dx \)
We use the double angle identities: \( \sin 2x = 2 \sin x \cos x \) and \( 1 + \cos 2x = 2 \cos^2 x \).
Substitute these into the expression inside \( \tan^{-1} \):
\( I = \int \tan^{-1}\left(\frac{2 \sin x \cos x}{2 \cos^2 x}\right) dx \)
Simplify the fraction:
\( I = \int \tan^{-1}\left(\frac{\sin x}{\cos x}\right) dx \)
\( I = \int \tan^{-1}(\tan x) dx \)
Since \( \tan^{-1}(\tan x) = x \):
\( I = \int x dx \)
Integrate \( x \):
\( I = \frac{x^2}{2} + C \)
In simple words: First, simplify the fraction inside the inverse tangent function using common trigonometry rules for double angles. This makes the fraction turn into just 'tan x'. Since inverse tangent of tan x is just x, we are left with a simple integral of x, which is easy to solve.

🎯 Exam Tip: Simplifying the argument of an inverse trigonometric function using identities is key to solving these integrals. Look for ways to reduce it to \( \tan^{-1}(\tan x) \), \( \sin^{-1}(\sin x) \), or \( \cos^{-1}(\cos x) \).

Question 11. Integrate the following: \( \cos^{-1}\left(\frac{1-\tan^2 x}{1+\tan^2 x}\right) \)
Answer: Let the integral be \( I \).
\( I = \int \cos^{-1}\left(\frac{1-\tan^2 x}{1+\tan^2 x}\right) dx \)
We use the double angle identity for cosine: \( \cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x} \).
Substitute this into the expression inside \( \cos^{-1} \):
\( I = \int \cos^{-1}(\cos 2x) dx \)
Since \( \cos^{-1}(\cos 2x) = 2x \):
\( I = \int 2x dx \)
Integrate \( 2x \):
\( I = 2 \cdot \frac{x^2}{2} + C \)
\( I = x^2 + C \)
In simple words: The expression inside the inverse cosine function is a special rule for 'cos 2x'. We replace it with 'cos 2x'. Since the inverse cosine of 'cos 2x' is just '2x', we simply need to integrate '2x'. This makes the problem much simpler.

🎯 Exam Tip: Recognizing standard trigonometric forms within inverse trigonometric functions is crucial. The expression \( \frac{1-\tan^2 x}{1+\tan^2 x} \) is a direct form for \( \cos 2x \).

Question 12. Integrate the following: \( \cos^{-1}(\sin x) \)
Answer: Let the integral be \( I \).
\( I = \int \cos^{-1}(\sin x) dx \)
We know that \( \sin x \) can be written as \( \cos \left(\frac{\pi}{2} - x\right) \).
Substitute this into the integral:
\( I = \int \cos^{-1}\left(\cos \left(\frac{\pi}{2} - x\right)\right) dx \)
Since \( \cos^{-1}(\cos A) = A \), here \( A = \frac{\pi}{2} - x \):
\( I = \int \left(\frac{\pi}{2} - x\right) dx \)
Now, integrate each term:
\( I = \frac{\pi}{2} x - \frac{x^2}{2} + C \)
In simple words: First, change 'sin x' into 'cos' of another angle, which is 'pi/2 minus x'. This way, we have 'inverse cosine of cos', which just cancels out to give 'pi/2 minus x'. Then, we integrate this simple expression to find the answer.

🎯 Exam Tip: When dealing with inverse trigonometric functions involving different functions (like \( \cos^{-1}(\sin x) \)), convert one function to match the other using angle complementary identities (e.g., \( \sin x = \cos(\frac{\pi}{2} - x) \)).

Question 13. If \( f'(x) = 3x^2 - \frac{2}{x^3} \) and \( f(1) = 0 \), find \( f(x) \).
Answer: We are given the derivative of a function: \( f'(x) = 3x^2 - \frac{2}{x^3} \).
To find \( f(x) \), we need to integrate \( f'(x) \) with respect to \( x \).
\( f(x) = \int \left(3x^2 - 2x^{-3}\right) dx \)
Integrate each term using the power rule \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \):
\( f(x) = 3 \cdot \frac{x^{2+1}}{2+1} - 2 \cdot \frac{x^{-3+1}}{-3+1} + C \)
\( f(x) = 3 \cdot \frac{x^3}{3} - 2 \cdot \frac{x^{-2}}{-2} + C \)
\( f(x) = x^3 + x^{-2} + C \)
\( f(x) = x^3 + \frac{1}{x^2} + C \)
We are given that \( f(1) = 0 \). We use this information to find the value of \( C \).
Substitute \( x = 1 \) and \( f(x) = 0 \) into the equation for \( f(x) \):
\( 0 = (1)^3 + \frac{1}{(1)^2} + C \)
\( 0 = 1 + 1 + C \)
\( 0 = 2 + C \)
\( C = -2 \)
Now, substitute the value of \( C \) back into the equation for \( f(x) \):
\( f(x) = x^3 + \frac{1}{x^2} - 2 \)
In simple words: To find the original function from its derivative, we perform integration. After integrating, we get a constant 'C'. We use the given condition, \( f(1)=0 \), to find the exact value of 'C'. Once 'C' is known, we write down the full function.

🎯 Exam Tip: Remember to always include the constant of integration \( C \) when finding \( f(x) \) from \( f'(x) \). The given boundary condition (like \( f(1)=0 \)) is used to determine the specific value of \( C \) for that particular function.

Question 14. If \( f'(x) = a \sin x + b \cos x \) and \( f'(0) = 4, f(0) = 3, f\left(\frac{\pi}{2}\right) = 5, \) find \( f(x) \).
Answer: We are given \( f'(x) = a \sin x + b \cos x \).
First, let's use the condition \( f'(0) = 4 \):
Substitute \( x = 0 \) into \( f'(x) \):
\( f'(0) = a \sin(0) + b \cos(0) \)
\( 4 = a(0) + b(1) \)
\( 4 = b \)
So, we have found that \( b = 4 \).
Now, we integrate \( f'(x) \) to find \( f(x) \):
\( f(x) = \int (a \sin x + b \cos x) dx \)
\( f(x) = -a \cos x + b \sin x + C \) (This is equation 2)
Next, let's use the condition \( f(0) = 3 \). Substitute \( x = 0 \) into the equation for \( f(x) \):
\( f(0) = -a \cos(0) + b \sin(0) + C \)
\( 3 = -a(1) + b(0) + C \)
\( 3 = -a + C \) (This is equation 3)
Finally, let's use the condition \( f\left(\frac{\pi}{2}\right) = 5 \). Substitute \( x = \frac{\pi}{2} \) into the equation for \( f(x) \):
\( f\left(\frac{\pi}{2}\right) = -a \cos\left(\frac{\pi}{2}\right) + b \sin\left(\frac{\pi}{2}\right) + C \)
\( 5 = -a(0) + b(1) + C \)
\( 5 = b + C \)
We already found \( b = 4 \). Substitute this into the equation above:
\( 5 = 4 + C \)
\( C = 1 \)
Now we have the values for \( b \) and \( C \). We can find \( a \) using equation 3:
\( 3 = -a + C \)
\( 3 = -a + 1 \)
\( a = 1 - 3 \)
\( a = -2 \)
Now we have all the constants: \( a = -2, b = 4, C = 1 \).
Substitute these values back into the equation for \( f(x) \):
\( f(x) = -(-2) \cos x + (4) \sin x + 1 \)
\( f(x) = 2 \cos x + 4 \sin x + 1 \)
In simple words: We are given the derivative and some values of the function. First, we use the derivative's value at \( x=0 \) to find one constant. Then we integrate the derivative to get the main function, which will have a new constant. We use the other given values of the function at different points to find the remaining constants. Finally, we put all the constants back into the function to get the complete answer.

🎯 Exam Tip: This type of problem involves multiple steps. Systematically use each given condition (for \( f'(x) \) and \( f(x) \)) one by one to find the unknown constants. Always list your known values clearly before substitution.