OP Malhotra Class 12 Maths Solutions Chapter 15 Indefinite Integral 3 Exercise 15 (G)

Question 1.
(i) \( \int \frac{dx}{4+5 \cos x} \)
(ii) \( \int \frac{d x}{12+12 \cos x} \)
(iii) \( \int \frac{d \theta}{4 \cos \theta-1} \)
(iv) \( \int \frac{d \theta}{1+2 \cos \theta} \)
Answer:
(i) To integrate \( \int \frac{dx}{4+5 \cos x} \), we use the substitution \( t = \tan \frac{x}{2} \).
\( \implies dx = \frac{2 dt}{1+t^2} \)
\( \implies \cos x = \frac{1-t^2}{1+t^2} \)
Substitute these into the integral:
\( I = \int \frac{\frac{2dt}{1+t^2}}{4+5 \left(\frac{1-t^2}{1+t^2}\right)} \)
\( \implies I = \int \frac{2dt}{4(1+t^2)+5(1-t^2)} \)
\( \implies I = \int \frac{2dt}{4+4t^2+5-5t^2} \)
\( \implies I = \int \frac{2dt}{9-t^2} \)
\( \implies I = 2 \int \frac{dt}{3^2-t^2} \)
Using the standard integral formula \( \int \frac{dx}{a^2-x^2} = \frac{1}{2a} \log \left|\frac{a+x}{a-x}\right| + C \):
\( \implies I = 2 \times \frac{1}{2 \times 3} \log \left|\frac{3+t}{3-t}\right| + C \)
\( \implies I = \frac{1}{3} \log \left|\frac{3+\tan \frac{x}{2}}{3-\tan \frac{x}{2}}\right| + C \)
(ii) To integrate \( \int \frac{dx}{12+12 \cos x} \), we use the identity \( 1+\cos x = 2 \cos^2 \frac{x}{2} \).
\( \implies 12+12 \cos x = 12(1+\cos x) = 12 \times 2 \cos^2 \frac{x}{2} = 24 \cos^2 \frac{x}{2} \)
So, the integral becomes:
\( I = \int \frac{dx}{24 \cos^2 \frac{x}{2}} \)
\( \implies I = \frac{1}{24} \int \sec^2 \frac{x}{2} dx \)
Now, we integrate using \( \int \sec^2(ax) dx = \frac{1}{a} \tan(ax) + C \):
\( \implies I = \frac{1}{24} \left( \frac{\tan \frac{x}{2}}{\frac{1}{2}} \right) + C \)
\( \implies I = \frac{1}{24} (2 \tan \frac{x}{2}) + C \)
\( \implies I = \frac{1}{12} \tan \frac{x}{2} + C \)
(iii) To integrate \( \int \frac{d \theta}{4 \cos \theta-1} \), we again use the substitution \( t = \tan \frac{\theta}{2} \).
\( \implies d\theta = \frac{2 d t}{1+t^2} \)
\( \implies \cos \theta = \frac{1-t^2}{1+t^2} \)
Substitute into the integral:
\( I = \int \frac{\frac{2dt}{1+t^2}}{4 \left(\frac{1-t^2}{1+t^2}\right)-1} \)
\( \implies I = \int \frac{2dt}{4(1-t^2)-(1+t^2)} \)
\( \implies I = \int \frac{2dt}{4-4t^2-1-t^2} \)
\( \implies I = \int \frac{2dt}{3-5t^2} \)
\( \implies I = \frac{2}{5} \int \frac{dt}{\frac{3}{5}-t^2} \)
\( \implies I = \frac{2}{5} \int \frac{dt}{\left(\sqrt{\frac{3}{5}}\right)^2-t^2} \)
Using the formula \( \int \frac{dx}{a^2-x^2} = \frac{1}{2a} \log \left|\frac{a+x}{a-x}\right| + C \), where \( a = \sqrt{\frac{3}{5}} \):
\( \implies I = \frac{2}{5} \times \frac{1}{2 \sqrt{\frac{3}{5}}} \log \left|\frac{\sqrt{\frac{3}{5}}+t}{\sqrt{\frac{3}{5}}-t}\right| + C \)
\( \implies I = \frac{1}{5 \sqrt{\frac{3}{5}}} \log \left|\frac{\sqrt{\frac{3}{5}}+t}{\sqrt{\frac{3}{5}}-t}\right| + C \)
\( \implies I = \frac{1}{\sqrt{15}} \log \left|\frac{\sqrt{3}+\sqrt{5}t}{\sqrt{3}-\sqrt{5}t}\right| + C \)
Substitute back \( t = \tan \frac{\theta}{2} \):
\( \implies I = \frac{1}{\sqrt{15}} \log \left|\frac{\sqrt{3}+\sqrt{5} \tan \frac{\theta}{2}}{\sqrt{3}-\sqrt{5} \tan \frac{\theta}{2}}\right| + C \)
(iv) To integrate \( \int \frac{d \theta}{1+2 \cos \theta} \), we use the substitution \( t = \tan \frac{\theta}{2} \).
\( \implies d\theta = \frac{2 d t}{1+t^2} \)
\( \implies \cos \theta = \frac{1-t^2}{1+t^2} \)
Substitute into the integral:
\( I = \int \frac{\frac{2dt}{1+t^2}}{1+2\left(\frac{1-t^2}{1+t^2}\right)} \)
\( \implies I = \int \frac{2dt}{(1+t^2)+2(1-t^2)} \)
\( \implies I = \int \frac{2dt}{1+t^2+2-2t^2} \)
\( \implies I = \int \frac{2dt}{3-t^2} \)
\( \implies I = 2 \int \frac{dt}{(\sqrt{3})^2-t^2} \)
Using the formula \( \int \frac{dx}{a^2-x^2} = \frac{1}{2a} \log \left|\frac{a+x}{a-x}\right| + C \):
\( \implies I = 2 \times \frac{1}{2\sqrt{3}} \log \left|\frac{\sqrt{3}+t}{\sqrt{3}-t}\right| + C \)
\( \implies I = \frac{1}{\sqrt{3}} \log \left|\frac{\sqrt{3}+\tan \frac{\theta}{2}}{\sqrt{3}-\tan \frac{\theta}{2}}\right| + C \)
In simple words: For integrals involving \( \sin x \) or \( \cos x \) in the denominator, a common method is to substitute \( t = \tan \frac{x}{2} \). This turns the integral into a rational function of \( t \), which can then be solved using standard formulas like those for \( \int \frac{dx}{a^2 \pm x^2} \).

🎯 Exam Tip: Remember the substitutions for \( dx \), \( \sin x \), and \( \cos x \) in terms of \( t = \tan \frac{x}{2} \) to simplify trigonometric integrals into algebraic forms quickly.

Question 2.
(i) \( \int \frac{d x}{4-3 \sin x} \)
(ii) \( \int \frac{d x}{4+5 \sin x} \)
(iii) \( \int \frac{d \theta}{1-2 \sin \theta} \)
(iv) \( \int \frac{d x}{1-3 \sin x} \)
Answer:
(i) To integrate \( \int \frac{d x}{4-3 \sin x} \), we use the substitution \( t = \tan \frac{x}{2} \).
\( \implies dx = \frac{2 d t}{1+t^2} \)
\( \implies \sin x = \frac{2t}{1+t^2} \)
Substitute these into the integral:
\( I = \int \frac{\frac{2dt}{1+t^2}}{4-3\left(\frac{2t}{1+t^2}\right)} \)
\( \implies I = \int \frac{2dt}{4(1+t^2)-6t} \)
\( \implies I = \int \frac{2dt}{4t^2-6t+4} \)
\( \implies I = \int \frac{dt}{2t^2-3t+2} \)
\( \implies I = \frac{1}{2} \int \frac{dt}{t^2-\frac{3}{2}t+1} \)
Complete the square for the denominator: \( t^2-\frac{3}{2}t+1 = \left(t-\frac{3}{4}\right)^2 + \left(\frac{\sqrt{7}}{4}\right)^2 \).
\( \implies I = \frac{1}{2} \int \frac{dt}{\left(t-\frac{3}{4}\right)^2 + \left(\frac{\sqrt{7}}{4}\right)^2} \)
Using the standard integral formula \( \int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \):
\( \implies I = \frac{1}{2} \times \frac{1}{\frac{\sqrt{7}}{4}} \tan^{-1}\left(\frac{t-\frac{3}{4}}{\frac{\sqrt{7}}{4}}\right) + C \)
\( \implies I = \frac{2}{\sqrt{7}} \tan^{-1}\left(\frac{4t-3}{\sqrt{7}}\right) + C \)
Substitute back \( t = \tan \frac{x}{2} \):
\( \implies I = \frac{2}{\sqrt{7}} \tan^{-1}\left(\frac{4 \tan \frac{x}{2}-3}{\sqrt{7}}\right) + C \)
(ii) To integrate \( \int \frac{d x}{4+5 \sin x} \), we use the substitution \( t = \tan \frac{x}{2} \).
\( \implies dx = \frac{2 d t}{1+t^2} \)
\( \implies \sin x = \frac{2t}{1+t^2} \)
Substitute these into the integral:
\( I = \int \frac{\frac{2dt}{1+t^2}}{4+5\left(\frac{2t}{1+t^2}\right)} \)
\( \implies I = \int \frac{2dt}{4(1+t^2)+10t} \)
\( \implies I = \int \frac{2dt}{4t^2+10t+4} \)
\( \implies I = \int \frac{dt}{2t^2+5t+2} \)
\( \implies I = \frac{1}{2} \int \frac{dt}{t^2+\frac{5}{2}t+1} \)
Complete the square for the denominator: \( t^2+\frac{5}{2}t+1 = \left(t+\frac{5}{4}\right)^2 - \left(\frac{3}{4}\right)^2 \).
\( \implies I = \frac{1}{2} \int \frac{dt}{\left(t+\frac{5}{4}\right)^2 - \left(\frac{3}{4}\right)^2} \)
Using the standard integral formula \( \int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log \left|\frac{x-a}{x+a}\right| + C \):
\( \implies I = \frac{1}{2} \times \frac{1}{2 \times \frac{3}{4}} \log \left|\frac{t+\frac{5}{4}-\frac{3}{4}}{t+\frac{5}{4}+\frac{3}{4}}\right| + C \)
\( \implies I = \frac{1}{3} \log \left|\frac{t+\frac{1}{2}}{t+2}\right| + C \)
Substitute back \( t = \tan \frac{x}{2} \):
\( \implies I = \frac{1}{3} \log \left|\frac{2 \tan \frac{x}{2}+1}{2 \tan \frac{x}{2}+4}\right| + C \)
(iii) To integrate \( \int \frac{d \theta}{1-2 \sin \theta} \), we use the substitution \( t = \tan \frac{\theta}{2} \).
\( \implies d\theta = \frac{2 d t}{1+t^2} \)
\( \implies \sin \theta = \frac{2t}{1+t^2} \)
Substitute into the integral:
\( I = \int \frac{\frac{2dt}{1+t^2}}{1-2\left(\frac{2t}{1+t^2}\right)} \)
\( \implies I = \int \frac{2dt}{1+t^2-4t} \)
\( \implies I = \int \frac{2dt}{t^2-4t+1} \)
Complete the square for the denominator: \( t^2-4t+1 = (t-2)^2 - (\sqrt{3})^2 \).
\( \implies I = 2 \int \frac{dt}{(t-2)^2-(\sqrt{3})^2} \)
Using the formula \( \int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log \left|\frac{x-a}{x+a}\right| + C \):
\( \implies I = 2 \times \frac{1}{2\sqrt{3}} \log \left|\frac{t-2-\sqrt{3}}{t-2+\sqrt{3}}\right| + C \)
\( \implies I = \frac{1}{\sqrt{3}} \log \left|\frac{\tan \frac{\theta}{2}-2-\sqrt{3}}{\tan \frac{\theta}{2}-2+\sqrt{3}}\right| + C \)
(iv) To integrate \( \int \frac{d x}{1-3 \sin x} \), we use the substitution \( t = \tan \frac{x}{2} \).
\( \implies dx = \frac{2 d t}{1+t^2} \)
\( \implies \sin x = \frac{2t}{1+t^2} \)
Substitute into the integral:
\( I = \int \frac{\frac{2dt}{1+t^2}}{1-3\left(\frac{2t}{1+t^2}\right)} \)
\( \implies I = \int \frac{2dt}{1+t^2-6t} \)
\( \implies I = 2 \int \frac{dt}{t^2-6t+1} \)
Complete the square for the denominator: \( t^2-6t+1 = (t-3)^2 - (\sqrt{8})^2 \).
\( \implies I = 2 \int \frac{dt}{(t-3)^2-(\sqrt{8})^2} \)
Using the formula \( \int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log \left|\frac{x-a}{x+a}\right| + C \):
\( \implies I = 2 \times \frac{1}{2\sqrt{8}} \log \left|\frac{t-3-\sqrt{8}}{t-3+\sqrt{8}}\right| + C \)
\( \implies I = \frac{1}{\sqrt{8}} \log \left|\frac{t-3-\sqrt{8}}{t-3+\sqrt{8}}\right| + C \)
\( \implies I = \frac{1}{2\sqrt{2}} \log \left|\frac{\tan \frac{x}{2}-3-2\sqrt{2}}{\tan \frac{x}{2}-3+2\sqrt{2}}\right| + C \)
In simple words: When you have integrals with \( \sin x \) or \( \cos x \) in the denominator, the trick is to use the half-angle tangent substitution. This converts trigonometric functions into algebraic expressions, making them easier to integrate using standard formulas.

🎯 Exam Tip: Always remember the conversion formulas for \( dx \), \( \sin x \), and \( \cos x \) in terms of \( t = \tan \frac{x}{2} \) and know how to complete the square for quadratic expressions in the denominator.

Question 3.
(i) \( \int \frac{d x}{2 \sin x-3 \cos x} \)
(ii) \( \int \frac{d x}{3 \cos x+4 \sin x} \)
(iii) \( \int \frac{d x}{\sin x-\cos x} \)
Answer:
(i) To integrate \( \int \frac{d x}{2 \sin x-3 \cos x} \), we use the substitution \( t = \tan \frac{x}{2} \).
\( \implies dx = \frac{2 d t}{1+t^2} \)
\( \implies \sin x = \frac{2t}{1+t^2} \)
\( \implies \cos x = \frac{1-t^2}{1+t^2} \)
Substitute these into the integral:
\( I = \int \frac{\frac{2dt}{1+t^2}}{2\left(\frac{2t}{1+t^2}\right)-3\left(\frac{1-t^2}{1+t^2}\right)} \)
\( \implies I = \int \frac{2dt}{4t-3(1-t^2)} \)
\( \implies I = \int \frac{2dt}{4t-3+3t^2} \)
\( \implies I = \int \frac{2dt}{3t^2+4t-3} \)
\( \implies I = \frac{2}{3} \int \frac{dt}{t^2+\frac{4}{3}t-1} \)
Complete the square for the denominator: \( t^2+\frac{4}{3}t-1 = \left(t+\frac{2}{3}\right)^2 - \left(\frac{\sqrt{13}}{3}\right)^2 \).
\( \implies I = \frac{2}{3} \int \frac{dt}{\left(t+\frac{2}{3}\right)^2 - \left(\frac{\sqrt{13}}{3}\right)^2} \)
Using the standard integral formula \( \int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log \left|\frac{x-a}{x+a}\right| + C \):
\( \implies I = \frac{2}{3} \times \frac{1}{2 \times \frac{\sqrt{13}}{3}} \log \left|\frac{t+\frac{2}{3}-\frac{\sqrt{13}}{3}}{t+\frac{2}{3}+\frac{\sqrt{13}}{3}}\right| + C \)
\( \implies I = \frac{1}{\sqrt{13}} \log \left|\frac{3t+2-\sqrt{13}}{3t+2+\sqrt{13}}\right| + C \)
Substitute back \( t = \tan \frac{x}{2} \):
\( \implies I = \frac{1}{\sqrt{13}} \log \left|\frac{3 \tan \frac{x}{2}+2-\sqrt{13}}{3 \tan \frac{x}{2}+2+\sqrt{13}}\right| + C \)
(ii) To integrate \( \int \frac{d x}{3 \cos x+4 \sin x} \), we use the substitution \( t = \tan \frac{x}{2} \).
\( \implies dx = \frac{2 d t}{1+t^2} \)
\( \implies \sin x = \frac{2t}{1+t^2} \)
\( \implies \cos x = \frac{1-t^2}{1+t^2} \)
Substitute these into the integral:
\( I = \int \frac{\frac{2dt}{1+t^2}}{3\left(\frac{1-t^2}{1+t^2}\right)+4\left(\frac{2t}{1+t^2}\right)} \)
\( \implies I = \int \frac{2dt}{3(1-t^2)+8t} \)
\( \implies I = \int \frac{2dt}{3-3t^2+8t} \)
\( \implies I = \int \frac{2dt}{-3t^2+8t+3} \)
\( \implies I = -\frac{2}{3} \int \frac{dt}{t^2-\frac{8}{3}t-1} \)
Complete the square for the denominator: \( t^2-\frac{8}{3}t-1 = \left(t-\frac{4}{3}\right)^2 - \left(\frac{5}{3}\right)^2 \).
\( \implies I = -\frac{2}{3} \int \frac{dt}{\left(t-\frac{4}{3}\right)^2 - \left(\frac{5}{3}\right)^2} \)
Using the standard integral formula \( \int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log \left|\frac{x-a}{x+a}\right| + C \):
\( \implies I = -\frac{2}{3} \times \frac{1}{2 \times \frac{5}{3}} \log \left|\frac{t-\frac{4}{3}-\frac{5}{3}}{t-\frac{4}{3}+\frac{5}{3}}\right| + C \)
\( \implies I = -\frac{1}{5} \log \left|\frac{t-3}{t+\frac{1}{3}}\right| + C \)
Substitute back \( t = \tan \frac{x}{2} \):
\( \implies I = -\frac{1}{5} \log \left|\frac{3 \tan \frac{x}{2}-9}{3 \tan \frac{x}{2}+1}\right| + C \)
(iii) To integrate \( \int \frac{d x}{\sin x-\cos x} \), we use the substitution \( t = \tan \frac{x}{2} \).
\( \implies dx = \frac{2 d t}{1+t^2} \)
\( \implies \sin x = \frac{2t}{1+t^2} \)
\( \implies \cos x = \frac{1-t^2}{1+t^2} \)
Substitute these into the integral:
\( I = \int \frac{\frac{2dt}{1+t^2}}{\frac{2t}{1+t^2}-\frac{1-t^2}{1+t^2}} \)
\( \implies I = \int \frac{2dt}{2t-(1-t^2)} \)
\( \implies I = \int \frac{2dt}{t^2+2t-1} \)
Complete the square for the denominator: \( t^2+2t-1 = (t+1)^2 - (\sqrt{2})^2 \).
\( \implies I = 2 \int \frac{dt}{(t+1)^2-(\sqrt{2})^2} \)
Using the standard integral formula \( \int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log \left|\frac{x-a}{x+a}\right| + C \):
\( \implies I = 2 \times \frac{1}{2\sqrt{2}} \log \left|\frac{t+1-\sqrt{2}}{t+1+\sqrt{2}}\right| + C \)
\( \implies I = \frac{1}{\sqrt{2}} \log \left|\frac{\tan \frac{x}{2}+1-\sqrt{2}}{\tan \frac{x}{2}+1+\sqrt{2}}\right| + C \)
In simple words: Integrals of the form \( \frac{1}{a \sin x + b \cos x} \) can be solved by converting \( \sin x \) and \( \cos x \) to half-angle tangent forms. This changes the problem into integrating a rational function, which often involves completing the square and using logarithmic formulas.

🎯 Exam Tip: When the denominator contains both \( \sin x \) and \( \cos x \), the \( t = \tan \frac{x}{2} \) substitution is usually the most effective approach. Be careful with algebraic manipulations and completing the square steps.

Question 4.
(i) \( \int \frac{d x}{3+2 \sin x+\cos x} \)
(ii) \( \int \frac{d x}{1-\sin x+\cos x} \)
(iii) \( \int \frac{d x}{1+\sin x+\cos x} \)
Answer:
(i) To integrate \( \int \frac{d x}{3+2 \sin x+\cos x} \), we use the substitution \( t = \tan \frac{x}{2} \).
\( \implies dx = \frac{2 d t}{1+t^2} \)
\( \implies \sin x = \frac{2t}{1+t^2} \)
\( \implies \cos x = \frac{1-t^2}{1+t^2} \)
Substitute these into the integral:
\( I = \int \frac{\frac{2dt}{1+t^2}}{3+2\left(\frac{2t}{1+t^2}\right)+\left(\frac{1-t^2}{1+t^2}\right)} \)
\( \implies I = \int \frac{2dt}{3(1+t^2)+4t+(1-t^2)} \)
\( \implies I = \int \frac{2dt}{3+3t^2+4t+1-t^2} \)
\( \implies I = \int \frac{2dt}{2t^2+4t+4} \)
\( \implies I = \int \frac{dt}{t^2+2t+2} \)
Complete the square for the denominator: \( t^2+2t+2 = (t+1)^2 + 1^2 \).
\( \implies I = \int \frac{dt}{(t+1)^2+1^2} \)
Using the standard integral formula \( \int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \):
\( \implies I = \frac{1}{1} \tan^{-1}\left(\frac{t+1}{1}\right) + C = \tan^{-1}(t+1) + C \)
Substitute back \( t = \tan \frac{x}{2} \):
\( \implies I = \tan^{-1}(1 + \tan \frac{x}{2}) + C \)
(ii) To integrate \( \int \frac{d x}{1-\sin x+\cos x} \), we use the substitution \( t = \tan \frac{x}{2} \).
\( \implies dx = \frac{2 d t}{1+t^2} \)
\( \implies \sin x = \frac{2t}{1+t^2} \)
\( \implies \cos x = \frac{1-t^2}{1+t^2} \)
Substitute these into the integral:
\( I = \int \frac{\frac{2dt}{1+t^2}}{1-\left(\frac{2t}{1+t^2}\right)+\left(\frac{1-t^2}{1+t^2}\right)} \)
\( \implies I = \int \frac{2dt}{(1+t^2)-2t+(1-t^2)} \)
\( \implies I = \int \frac{2dt}{1+t^2-2t+1-t^2} \)
\( \implies I = \int \frac{2dt}{2-2t} \)
\( \implies I = \int \frac{dt}{1-t} \)
\( \implies I = -\log |1-t| + C \)
Substitute back \( t = \tan \frac{x}{2} \):
\( \implies I = -\log |1 - \tan \frac{x}{2}| + C \)
(iii) To integrate \( \int \frac{d x}{1+\sin x+\cos x} \), we use the substitution \( t = \tan \frac{x}{2} \).
\( \implies dx = \frac{2 d t}{1+t^2} \)
\( \implies \sin x = \frac{2t}{1+t^2} \)
\( \implies \cos x = \frac{1-t^2}{1+t^2} \)
Substitute these into the integral:
\( I = \int \frac{\frac{2dt}{1+t^2}}{1+\left(\frac{2t}{1+t^2}\right)+\left(\frac{1-t^2}{1+t^2}\right)} \)
\( \implies I = \int \frac{2dt}{(1+t^2)+2t+(1-t^2)} \)
\( \implies I = \int \frac{2dt}{1+t^2+2t+1-t^2} \)
\( \implies I = \int \frac{2dt}{2+2t} \)
\( \implies I = \int \frac{dt}{1+t} \)
\( \implies I = \log |1+t| + C \)
Substitute back \( t = \tan \frac{x}{2} \):
\( \implies I = \log |1 + \tan \frac{x}{2}| + C \)
In simple words: When an integral has a combination of a constant, \( \sin x \), and \( \cos x \) in the denominator, the \( t = \tan \frac{x}{2} \) substitution simplifies it into a rational function. You can then use methods like completing the square and standard integration formulas.

🎯 Exam Tip: Pay close attention to the signs in the denominator after substitution, as they determine whether you use \( \int \frac{dx}{x^2+a^2} \) or \( \int \frac{dx}{a^2-x^2} \) or \( \int \frac{dx}{x^2-a^2} \).

Question 5.
(i) \( \int \frac{2 \sin \theta+\cos \theta}{7 \sin \theta-5 \cos \theta} d \theta \)
(ii) \( \int \frac{\cos x+3 \sin x-5}{3(1-\sin x)-\cos x} d x \)
Answer:
(i) To integrate \( \int \frac{2 \sin \theta+\cos \theta}{7 \sin \theta-5 \cos \theta} d \theta \), we express the numerator as a linear combination of the denominator and its derivative.
Let \( 2 \sin \theta + \cos \theta = L(7 \sin \theta - 5 \cos \theta) + M \frac{d}{d\theta}(7 \sin \theta - 5 \cos \theta) \).
\( \implies 2 \sin \theta + \cos \theta = L(7 \sin \theta - 5 \cos \theta) + M(7 \cos \theta + 5 \sin \theta) \)
\( \implies 2 \sin \theta + \cos \theta = (7L+5M) \sin \theta + (-5L+7M) \cos \theta \)
Comparing coefficients:
\( 7L+5M = 2 \) ...(1)
\( -5L+7M = 1 \) ...(2)
Solving these equations gives \( L = \frac{9}{74} \) and \( M = \frac{17}{74} \).
So, \( 2 \sin \theta + \cos \theta = \frac{9}{74}(7 \sin \theta - 5 \cos \theta) + \frac{17}{74}(7 \cos \theta + 5 \sin \theta) \).
The integral becomes:
\( I = \int \left( \frac{9}{74} + \frac{17}{74} \frac{7 \cos \theta + 5 \sin \theta}{7 \sin \theta - 5 \cos \theta} \right) d \theta \)
\( \implies I = \frac{9}{74} \int d\theta + \frac{17}{74} \int \frac{7 \cos \theta + 5 \sin \theta}{7 \sin \theta - 5 \cos \theta} d \theta \)
The second integral is of the form \( \int \frac{f'(\theta)}{f(\theta)} d\theta = \log |f(\theta)| + C \).
\( \implies I = \frac{9}{74} \theta + \frac{17}{74} \log |7 \sin \theta - 5 \cos \theta| + C \)
(ii) To integrate \( \int \frac{\cos x+3 \sin x-5}{3(1-\sin x)-\cos x} d x \), we express the numerator in terms of the denominator and its derivative.
Let \( \text{Numerator} = L(\text{Denominator}) + M \frac{d}{dx}(\text{Denominator}) + N \).
\( \cos x + 3 \sin x - 5 = L(3 - 3 \sin x - \cos x) + M(-3 \cos x + \sin x) + N \)
Comparing coefficients, we get:
\( 3 = 3L+M \)
\( 1 = -L-3M \)
\( -5 = 3L+N \)
Solving these equations, we find \( L = \frac{5}{4} \), \( M = -\frac{3}{4} \), and \( N = -\frac{35}{4} \).
So, \( I = \int \left( \frac{5}{4} - \frac{3}{4} \frac{-3 \cos x + \sin x}{3-3 \sin x - \cos x} - \frac{35}{4} \frac{1}{3-3 \sin x - \cos x} \right) d x \)
\( \implies I = \frac{5}{4} x - \frac{3}{4} \log |3-3 \sin x - \cos x| - \frac{35}{4} \int \frac{dx}{3-3 \sin x - \cos x} \)
Let \( I_1 = \int \frac{dx}{3-3 \sin x - \cos x} \). We use the substitution \( t = \tan \frac{x}{2} \).
\( \implies dx = \frac{2dt}{1+t^2} \), \( \sin x = \frac{2t}{1+t^2} \), \( \cos x = \frac{1-t^2}{1+t^2} \)
\( I_1 = \int \frac{\frac{2dt}{1+t^2}}{3-3\left(\frac{2t}{1+t^2}\right)-\left(\frac{1-t^2}{1+t^2}\right)} = \int \frac{2dt}{3(1+t^2)-6t-(1-t^2)} \)
\( \implies I_1 = \int \frac{2dt}{4t^2-6t+2} = \int \frac{dt}{2t^2-3t+1} \)
\( \implies I_1 = \frac{1}{2} \int \frac{dt}{t^2-\frac{3}{2}t+\frac{1}{2}} \)
Completing the square: \( t^2-\frac{3}{2}t+\frac{1}{2} = \left(t-\frac{3}{4}\right)^2 - \left(\frac{1}{4}\right)^2 \).
\( \implies I_1 = \frac{1}{2} \int \frac{dt}{\left(t-\frac{3}{4}\right)^2 - \left(\frac{1}{4}\right)^2} \)
Using \( \int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log \left|\frac{x-a}{x+a}\right| + C \):
\( \implies I_1 = \frac{1}{2} \times \frac{1}{2 \times \frac{1}{4}} \log \left|\frac{t-\frac{3}{4}-\frac{1}{4}}{t-\frac{3}{4}+\frac{1}{4}}\right| + C = \log \left|\frac{2t-2}{2t-1}\right| + C \)
Substitute back \( t = \tan \frac{x}{2} \):
\( \implies I_1 = \log \left|\frac{2 \tan \frac{x}{2}-2}{2 \tan \frac{x}{2}-1}\right| + C \).
Finally, substitute \( I_1 \) back into \( I \):
\( I = \frac{5}{4} x - \frac{3}{4} \log |3-3 \sin x - \cos x| - \frac{35}{4} \log \left|\frac{2 \tan \frac{x}{2}-2}{2 \tan \frac{x}{2}-1}\right| + C \)
In simple words: When the numerator is a linear combination of \( \sin x \) and \( \cos x \) and the denominator is also a linear combination, we can express the numerator using the denominator and its derivative. For more complex cases with a constant term, an additional constant 'N' is added. The remaining integral usually requires the \( t = \tan \frac{x}{2} \) substitution.

🎯 Exam Tip: This method is crucial for integrals of the form \( \int \frac{a \sin x + b \cos x + c}{p \sin x + q \cos x + r} dx \). Accurately setting up and solving the system of linear equations for L, M, and N is the most important step.

Question 6.
(i) \( \int \frac{d x}{4 \sin ^2 x+5 \cos ^2 x} \)
(ii) \( \int \frac{d x}{2 \cos ^2 x+\sin ^2 x} \)
(iii) \( \int \frac{d x}{1+3 \sin ^2 x} \)
(iv) \( \int \frac{d x}{\cos 2 x+3 \sin ^2 x} \)
Answer:
(i) To integrate \( \int \frac{d x}{4 \sin ^2 x+5 \cos ^2 x} \), we divide the numerator and denominator by \( \cos^2 x \).
\( \implies I = \int \frac{\sec^2 x dx}{4 \tan^2 x+5} \)
Let \( t = \tan x \), so \( dt = \sec^2 x dx \).
\( \implies I = \int \frac{dt}{4t^2+5} \)
\( \implies I = \frac{1}{4} \int \frac{dt}{t^2+\frac{5}{4}} \)
\( \implies I = \frac{1}{4} \int \frac{dt}{t^2+\left(\frac{\sqrt{5}}{2}\right)^2} \)
Using the standard integral formula \( \int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \):
\( \implies I = \frac{1}{4} \times \frac{1}{\frac{\sqrt{5}}{2}} \tan^{-1}\left(\frac{t}{\frac{\sqrt{5}}{2}}\right) + C \)
\( \implies I = \frac{1}{2\sqrt{5}} \tan^{-1}\left(\frac{2 \tan x}{\sqrt{5}}\right) + C \)
(ii) To integrate \( \int \frac{d x}{2 \cos ^2 x+\sin ^2 x} \), we divide the numerator and denominator by \( \cos^2 x \).
\( \implies I = \int \frac{\sec^2 x dx}{2 + \tan^2 x} \)
Let \( t = \tan x \), so \( dt = \sec^2 x dx \).
\( \implies I = \int \frac{dt}{t^2+(\sqrt{2})^2} \)
Using the formula \( \int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \):
\( \implies I = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{t}{\sqrt{2}}\right) + C \)
\( \implies I = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{\tan x}{\sqrt{2}}\right) + C \)
(iii) To integrate \( \int \frac{d x}{1+3 \sin ^2 x} \), we divide the numerator and denominator by \( \cos^2 x \).
\( \implies I = \int \frac{\sec^2 x dx}{\sec^2 x + 3 \tan^2 x} \)
Using the identity \( \sec^2 x = 1+\tan^2 x \):
\( \implies I = \int \frac{\sec^2 x dx}{1+\tan^2 x + 3 \tan^2 x} = \int \frac{\sec^2 x dx}{1+4 \tan^2 x} \)
Let \( t = \tan x \), so \( dt = \sec^2 x dx \).
\( \implies I = \int \frac{dt}{1+4t^2} \)
\( \implies I = \frac{1}{4} \int \frac{dt}{t^2+\left(\frac{1}{2}\right)^2} \)
Using the formula \( \int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \):
\( \implies I = \frac{1}{4} \times \frac{1}{\frac{1}{2}} \tan^{-1}\left(\frac{t}{\frac{1}{2}}\right) + C \)
\( \implies I = \frac{1}{2} \tan^{-1}(2 \tan x) + C \)
(iv) To integrate \( \int \frac{d x}{\cos 2 x+3 \sin ^2 x} \), we use the identity \( \cos 2x = \cos^2 x - \sin^2 x \).
\( \implies I = \int \frac{d x}{\cos^2 x - \sin^2 x + 3 \sin^2 x} = \int \frac{d x}{\cos^2 x + 2 \sin^2 x} \)
Now, divide the numerator and denominator by \( \cos^2 x \).
\( \implies I = \int \frac{\sec^2 x dx}{1 + 2 \tan^2 x} \)
Let \( t = \tan x \), so \( dt = \sec^2 x dx \).
\( \implies I = \int \frac{dt}{1+2t^2} \)
\( \implies I = \frac{1}{2} \int \frac{dt}{t^2+\left(\frac{1}{\sqrt{2}}\right)^2} \)
Using the formula \( \int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \):
\( \implies I = \frac{1}{2} \times \frac{1}{\frac{1}{\sqrt{2}}} \tan^{-1}\left(\frac{t}{\frac{1}{\sqrt{2}}}\right) + C \)
\( \implies I = \frac{1}{\sqrt{2}} \tan^{-1}(\sqrt{2} \tan x) + C \)
In simple words: For integrals where the denominator has only even powers of \( \sin x \) and \( \cos x \), or expressions that can be converted to \( \tan x \) and \( \sec^2 x \), divide the numerator and denominator by \( \cos^2 x \). Then, substitute \( t = \tan x \) to simplify the integral into a standard algebraic form.

🎯 Exam Tip: Always look for ways to convert trigonometric integrals into rational functions of \( \tan x \) by dividing by powers of \( \cos x \). This is a standard and very effective technique for these types of problems.

Question 6.
(v) \( \int \frac{d x}{9 \cos ^2 x+8 \sin ^2 x+3} \)
Answer: To solve this integral, we first divide the numerator and denominator by \( \cos^2 x \). This transforms the integral into terms of \( \tan x \) and \( \sec^2 x \).
So, we have:
\[ = \int \frac{\sec^2 x \, dx}{9+8 \tan^2 x+3 \sec^2 x} \]
Next, we replace \( \sec^2 x \) in the denominator with \( 1+\tan^2 x \):
\[ = \int \frac{\sec^2 x \, dx}{9+8 \tan^2 x+3 (1 + \tan^2 x)} \]
\[ = \int \frac{\sec^2 x \, dx}{9+8 \tan^2 x+3+3 \tan^2 x} \]
\[ = \int \frac{\sec^2 x \, dx}{12+11 \tan^2 x} \]
Now, we factor out 11 from the denominator:
\[ = \frac{1}{11} \int \frac{\sec^2 x \, dx}{\frac{12}{11}+\tan^2 x} \]
Let \( \tan x = t \). Then, \( \sec^2 x \, dx = dt \). Substituting these values into the integral:
\[ = \frac{1}{11} \int \frac{dt}{\frac{12}{11}+t^2} \]
We can write \( \frac{12}{11} \) as \( \left(\sqrt{\frac{12}{11}}\right)^2 \). This is in the form of \( \int \frac{dx}{a^2+x^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \), where \( a = \sqrt{\frac{12}{11}} \).
\[ = \frac{1}{11} \times \frac{1}{\sqrt{\frac{12}{11}}} \tan^{-1}\left(\frac{t}{\sqrt{\frac{12}{11}}}\right) + C \]
\[ = \frac{1}{11} \times \frac{\sqrt{11}}{\sqrt{12}} \tan^{-1}\left(\frac{t\sqrt{11}}{\sqrt{12}}\right) + C \]
Simplify the coefficients:
\[ = \frac{1}{\sqrt{11}\sqrt{12}} \tan^{-1}\left(\frac{t\sqrt{11}}{\sqrt{12}}\right) + C \]
Since \( \sqrt{11}\sqrt{12} = \sqrt{132} \), and substituting \( t = \tan x \):
\[ = \frac{1}{\sqrt{132}} \tan^{-1}\left(\frac{\tan x \sqrt{11}}{\sqrt{12}}\right) + C \]
\[ = \frac{1}{2\sqrt{33}} \tan^{-1}\left(\frac{\sqrt{11}\tan x}{2\sqrt{3}}\right) + C \] The integral is solved by using a substitution after making the denominator suitable for a standard integral form.
In simple words: First, change the integral so it only has tan x and sec²x. Then, replace sec²x to get everything in terms of tan x. After that, let t be tan x, and the integral becomes a common type that you can solve using a simple formula for tan inverse.

🎯 Exam Tip: When integrating expressions with \( \sin^2 x \) and \( \cos^2 x \) in the denominator, dividing both numerator and denominator by \( \cos^2 x \) is often the key first step to convert the expression into terms of \( \tan x \) and \( \sec^2 x \).

Question 6.
(vi) \( \int \frac{d x}{(2 \sin x+3 \cos x)^2} \)
Answer: To solve this integral, we begin by dividing both the numerator and denominator by \( \cos^2 x \). This helps to transform the expression into terms of \( \tan x \) and \( \sec^2 x \).
So, we get:
\[ = \int \frac{\frac{dx}{\cos^2 x}}{\left(\frac{2 \sin x+3 \cos x}{\cos x}\right)^2} \]
\[ = \int \frac{\sec^2 x \, dx}{(2 \tan x+3)^2} \]
Now, let \( 2 \tan x+3 = t \). To find \( dt \), we differentiate both sides with respect to x:
\( \frac{d}{dx}(2 \tan x+3) = \frac{dt}{dx} \)
\( 2 \sec^2 x = \frac{dt}{dx} \)
\( \sec^2 x \, dx = \frac{1}{2} dt \)
Substitute these into the integral:
\[ = \int \frac{\frac{1}{2} dt}{t^2} \]
\[ = \frac{1}{2} \int t^{-2} dt \]
Now, integrate \( t^{-2} \) using the power rule \( \int x^n dx = \frac{x^{n+1}}{n+1} \):
\[ = \frac{1}{2} \times \frac{t^{-2+1}}{-2+1} + C \]
\[ = \frac{1}{2} \times \frac{t^{-1}}{-1} + C \]
\[ = -\frac{1}{2t} + C \]
Finally, substitute back \( t = 2 \tan x+3 \):
\[ = -\frac{1}{2(2 \tan x+3)} + C \] This method uses a substitution after making the denominator suitable for integration.
In simple words: First, divide the top and bottom of the fraction by cos²x to make it easier to work with. This changes the terms into tan x and sec²x. Then, let the whole bottom part be 't' and find 'dt'. Replace these into the integral and solve it like a simple power rule problem.

🎯 Exam Tip: For integrals where the denominator is a squared linear combination of \( \sin x \) and \( \cos x \), dividing by \( \cos^2 x \) and then using a substitution for \( (a \tan x + b) \) simplifies the integral significantly.

Question 7.
(i) \( \int \frac{d x}{4+3 \sin 2 x} \)
(ii) \( \int \frac{d x}{\sin 2 x+4} \)
Answer:
(i) Let \( I = \int \frac{d x}{4+3 \sin 2 x} \).
We know that \( \sin 2x = 2 \sin x \cos x \). So, the integral becomes:
\[ = \int \frac{d x}{4+6 \sin x \cos x} \]
To solve this, we divide both the numerator and denominator by \( \cos^2 x \). This converts the expression into terms of \( \tan x \) and \( \sec^2 x \).
\[ = \int \frac{\frac{dx}{\cos^2 x}}{\frac{4}{\cos^2 x}+\frac{6 \sin x \cos x}{\cos^2 x}} \]
\[ = \int \frac{\sec^2 x \, dx}{4 \sec^2 x+6 \tan x} \]
Replace \( \sec^2 x \) in the denominator with \( 1+\tan^2 x \):
\[ = \int \frac{\sec^2 x \, dx}{4(1+\tan^2 x)+6 \tan x} \]
\[ = \int \frac{\sec^2 x \, dx}{4+4 \tan^2 x+6 \tan x} \]
Now, let \( \tan x = t \). Then, \( \sec^2 x \, dx = dt \). Substitute these into the integral:
\[ = \int \frac{dt}{4 t^2+6 t+4} \]
Factor out 4 from the denominator:
\[ = \frac{1}{4} \int \frac{dt}{t^2+\frac{6}{4} t+\frac{4}{4}} \]
\[ = \frac{1}{4} \int \frac{dt}{t^2+\frac{3}{2} t+1} \]
Complete the square in the denominator: \( t^2+\frac{3}{2} t+1 = t^2+2\left(\frac{3}{4}\right)t+\left(\frac{3}{4}\right)^2-\left(\frac{3}{4}\right)^2+1 \)
\[ = \left(t+\frac{3}{4}\right)^2 - \frac{9}{16} + 1 \]
\[ = \left(t+\frac{3}{4}\right)^2 + \frac{16-9}{16} \]
\[ = \left(t+\frac{3}{4}\right)^2 + \frac{7}{16} \]
So the integral becomes:
\[ = \frac{1}{4} \int \frac{dt}{\left(t+\frac{3}{4}\right)^2 + \left(\frac{\sqrt{7}}{4}\right)^2} \]
This is in the form \( \int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \), where \( x = t+\frac{3}{4} \) and \( a = \frac{\sqrt{7}}{4} \).
\[ = \frac{1}{4} \times \frac{1}{\frac{\sqrt{7}}{4}} \tan^{-1}\left(\frac{t+\frac{3}{4}}{\frac{\sqrt{7}}{4}}\right) + C \]
\[ = \frac{1}{4} \times \frac{4}{\sqrt{7}} \tan^{-1}\left(\frac{\frac{4t+3}{4}}{\frac{\sqrt{7}}{4}}\right) + C \]
\[ = \frac{1}{\sqrt{7}} \tan^{-1}\left(\frac{4t+3}{\sqrt{7}}\right) + C \]
Substitute back \( t = \tan x \):
\[ = \frac{1}{\sqrt{7}} \tan^{-1}\left(\frac{4 \tan x+3}{\sqrt{7}}\right) + C \] (ii) Let \( I = \int \frac{d x}{\sin 2 x+4} \).
We know that \( \sin 2x = 2 \sin x \cos x \). So, the integral becomes:
\[ = \int \frac{d x}{2 \sin x \cos x+4} \]
Divide both the numerator and denominator by \( \cos^2 x \):
\[ = \int \frac{\frac{dx}{\cos^2 x}}{\frac{2 \sin x \cos x}{\cos^2 x}+\frac{4}{\cos^2 x}} \]
\[ = \int \frac{\sec^2 x \, dx}{2 \tan x+4 \sec^2 x} \]
Replace \( \sec^2 x \) in the denominator with \( 1+\tan^2 x \):
\[ = \int \frac{\sec^2 x \, dx}{2 \tan x+4(1+\tan^2 x)} \]
\[ = \int \frac{\sec^2 x \, dx}{2 \tan x+4+4 \tan^2 x} \]
Now, let \( \tan x = t \). Then, \( \sec^2 x \, dx = dt \). Substitute these into the integral:
\[ = \int \frac{dt}{4 t^2+2 t+4} \]
Factor out 4 from the denominator:
\[ = \frac{1}{4} \int \frac{dt}{t^2+\frac{2}{4} t+\frac{4}{4}} \]
\[ = \frac{1}{4} \int \frac{dt}{t^2+\frac{1}{2} t+1} \]
Complete the square in the denominator: \( t^2+\frac{1}{2} t+1 = t^2+2\left(\frac{1}{4}\right)t+\left(\frac{1}{4}\right)^2-\left(\frac{1}{4}\right)^2+1 \)
\[ = \left(t+\frac{1}{4}\right)^2 - \frac{1}{16} + 1 \]
\[ = \left(t+\frac{1}{4}\right)^2 + \frac{16-1}{16} \]
\[ = \left(t+\frac{1}{4}\right)^2 + \frac{15}{16} \]
So the integral becomes:
\[ = \frac{1}{4} \int \frac{dt}{\left(t+\frac{1}{4}\right)^2 + \left(\frac{\sqrt{15}}{4}\right)^2} \]
This is in the form \( \int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \), where \( x = t+\frac{1}{4} \) and \( a = \frac{\sqrt{15}}{4} \).
\[ = \frac{1}{4} \times \frac{1}{\frac{\sqrt{15}}{4}} \tan^{-1}\left(\frac{t+\frac{1}{4}}{\frac{\sqrt{15}}{4}}\right) + C \]
\[ = \frac{1}{4} \times \frac{4}{\sqrt{15}} \tan^{-1}\left(\frac{\frac{4t+1}{4}}{\frac{\sqrt{15}}{4}}\right) + C \]
\[ = \frac{1}{\sqrt{15}} \tan^{-1}\left(\frac{4t+1}{\sqrt{15}}\right) + C \]
Substitute back \( t = \tan x \):
\[ = \frac{1}{\sqrt{15}} \tan^{-1}\left(\frac{4 \tan x+1}{\sqrt{15}}\right) + C \] To solve integrals like these, we convert them into standard forms using trigonometric identities and substitution. This involves making the denominator a perfect square plus a constant.
In simple words: For both parts, change sin 2x to 2 sin x cos x. Then, divide everything by cos²x to get tan x and sec²x. Let tan x be 't', so sec²x dx becomes 'dt'. This makes the integral look like a simple fraction with t² and a constant, which you can solve using the tan inverse formula.

🎯 Exam Tip: Remember to express \( \sin 2x \) in terms of \( \sin x \cos x \) and then divide by \( \cos^2 x \) to apply the substitution \( t = \tan x \). Completing the square in the denominator is a crucial step for these types of integrals.

Question 8. \( \int \frac{\sin 2 x}{\left(\sin ^4 x+\cos ^4x\right)} d x \)
Answer: Let \( I = \int \frac{\sin 2 x}{\left(\sin ^4 x+\cos ^4x\right)} d x \).
First, we use the identity \( \sin 2x = 2 \sin x \cos x \):
\[ I = \int \frac{2 \sin x \cos x}{\sin ^4 x+\cos ^4x} d x \]
To simplify the denominator, divide both the numerator and denominator by \( \cos^4 x \). This will convert the expression into terms of \( \tan x \) and \( \sec x \).
\[ I = \int \frac{\frac{2 \sin x \cos x}{\cos^4 x}}{\frac{\sin ^4 x}{\cos^4 x}+\frac{\cos ^4x}{\cos^4 x}} d x \]
\[ I = \int \frac{2 \frac{\sin x}{\cos x} \frac{1}{\cos^2 x}}{\tan ^4 x+1} d x \]
\[ I = \int \frac{2 \tan x \sec^2 x}{\tan ^4 x+1} d x \]
Now, let \( \tan^2 x = t \). To find \( dt \), we differentiate both sides with respect to x:
\( \frac{d}{dx}(\tan^2 x) = \frac{dt}{dx} \)
Using the chain rule, \( 2 \tan x \cdot \sec^2 x = \frac{dt}{dx} \)
So, \( 2 \tan x \sec^2 x \, dx = dt \).
Substitute these into the integral:
\[ I = \int \frac{dt}{t^2+1} \]
This is a standard integral form: \( \int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \), where \( a=1 \).
\[ I = \frac{1}{1} \tan^{-1}\left(\frac{t}{1}\right) + C \]
\[ I = \tan^{-1}(t) + C \]
Finally, substitute back \( t = \tan^2 x \):
\[ I = \tan^{-1}(\tan^2 x) + C \] The solution uses trigonometric identities and a clever substitution to simplify the integral into a basic form.
In simple words: First, change sin 2x to 2 sin x cos x. Then, divide the top and bottom of the fraction by cos⁴x. This helps turn everything into tan x and sec²x. After that, let tan²x be 't', which makes the top part of the fraction turn into 'dt'. The integral then becomes very simple, like finding the tan inverse of 't'.

🎯 Exam Tip: For integrals involving \( \sin^n x \) and \( \cos^n x \) in the denominator and \( \sin 2x \) in the numerator, dividing by the highest power of \( \cos x \) in the denominator (like \( \cos^4 x \) here) often leads to a substitution involving \( \tan^2 x \).