PYQs Homework Notes Mocks Board Daily Homework MHT-CET Maths Question type Integral of 1/(a + b cos 2x) or 1/(a + b sin^2 x): divide by cos^2 x and substitute t = tan x This type of question, asked 5 times: 2 May 2023 S1, 15 May 2023 S1, 25 Apr 2025 S1, 19 Apr 2026 S1, 13 May 2026 S1. All 5 questions below, oldest first.
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Question 1 2 May 2023 S1
147 ∫ d x 3 − 2 cos 2 x = tan − 1 ( f ( x ) ) 5 + c \displaystyle\int \dfrac{dx}{3 - 2\cos 2x} = \dfrac{\tan^{-1}(f(x))}{\sqrt{5}} + c ∫ 3 − 2 cos 2 x d x = 5 tan − 1 ( f ( x )) + c , (where c c c is constant of integration), then f ( π / 4 ) f(\pi/4) f ( π /4 ) has the value A. B. C. D. 1 5 \dfrac{1}{\sqrt{5}} 5 1 Tap an option to check your answer.
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Question 2 15 May 2023 S1
105 If ∫ d x 1 + 3 sin 2 x = 1 2 tan − 1 ( f ( x ) ) + c \int \frac{dx}{1+3\sin^{2}x} = \frac{1}{2}\tan^{-1}(f(x))+c ∫ 1 + 3 s i n 2 x d x = 2 1 tan − 1 ( f ( x )) + c , where c is a constant of integration, then f ( x ) f(x) f ( x ) is equal to A. B. C. D. Tap an option to check your answer.
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Question 3 25 Apr 2025 S1
122 ∫ d x 3 cos 2 x + 5 \int\frac{dx}{3\cos2x+ 5} ∫ 3 c o s 2 x + 5 d x equals A. 1 2 tan − 1 ( tan x ) + c \frac{1}{2}\tan^{- 1}(\tan x) + c 2 1 tan − 1 ( tan x ) + c , where c is the constant of integration. B. 1 2 tan − 1 ( tan x 2 ) + c \frac{1}{2}\tan^{- 1}\left( \frac{\tan x}{2} \right) + c 2 1 tan − 1 ( 2 t a n x ) + c , where c is the constant of integration. C. 1 4 tan − 1 ( 1 2 tan x ) + c \frac{1}{4}\tan^{- 1}\left( \frac{1}{2}\tan x \right) + c 4 1 tan − 1 ( 2 1 tan x ) + c , where c is the constant of integration. D. 1 4 tan − 1 ( tan x ) + c \frac{1}{4}\tan^{- 1}(\tan x) + c 4 1 tan − 1 ( tan x ) + c , where c is the constant of integration. Tap an option to check your answer.
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Question 4 19 Apr 2026 S1
78 The value of integral ∫ d x sin 2 x + tan 2 x \displaystyle\int\dfrac{dx}{\sin^2 x + \tan^2 x} ∫ sin 2 x + tan 2 x d x is... A. − 1 2 tan x + 1 2 2 tan − 1 ( tan x 2 ) + c \dfrac{-1}{2\tan x} + \dfrac{1}{2\sqrt{2}}\tan^{-1}\left(\dfrac{\tan x}{\sqrt{2}}\right) + c 2 tan x − 1 + 2 2 1 tan − 1 ( 2 tan x ) + c B. − 1 2 tan x − 1 2 2 tan − 1 ( tan x 2 ) + c \dfrac{-1}{2\tan x} - \dfrac{1}{2\sqrt{2}}\tan^{-1}\left(\dfrac{\tan x}{\sqrt{2}}\right) + c 2 tan x − 1 − 2 2 1 tan − 1 ( 2 tan x ) + c C. 1 2 tan x − 1 2 2 tan − 1 ( tan x 2 ) + c \dfrac{1}{2\tan x} - \dfrac{1}{2\sqrt{2}}\tan^{-1}\left(\dfrac{\tan x}{\sqrt{2}}\right) + c 2 tan x 1 − 2 2 1 tan − 1 ( 2 tan x ) + c D. 1 2 tan x + 1 2 2 tan − 1 ( tan x 2 ) + c \dfrac{1}{2\tan x} + \dfrac{1}{2\sqrt{2}}\tan^{-1}\left(\dfrac{\tan x}{\sqrt{2}}\right) + c 2 tan x 1 + 2 2 1 tan − 1 ( 2 tan x ) + c Tap an option to check your answer.
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Question 5 13 May 2026 S1
79 The value of the integral ∫ d x 3 + sin 2 x + sin 2 x = … \displaystyle\int\dfrac{dx}{3 + \sin 2x + \sin^2 x} = \ldots ∫ 3 + sin 2 x + sin 2 x d x = … A. 1 13 tan − 1 ( 4 tan x + 1 13 ) + c \dfrac{1}{\sqrt{13}}\tan^{-1}\left(\dfrac{4\tan x + 1}{\sqrt{13}}\right) + c 13 1 tan − 1 ( 13 4 tan x + 1 ) + c B. 2 11 tan − 1 ( 4 tan x + 2 11 ) + c \dfrac{2}{\sqrt{11}}\tan^{-1}\left(\dfrac{4\tan x + 2}{\sqrt{11}}\right) + c 11 2 tan − 1 ( 11 4 tan x + 2 ) + c C. 1 11 tan − 1 ( 2 tan x + 1 11 ) + c \dfrac{1}{11}\tan^{-1}\left(\dfrac{2\tan x + 1}{11}\right) + c 11 1 tan − 1 ( 11 2 tan x + 1 ) + c D. 1 11 tan − 1 ( 4 tan x + 1 11 ) + c \dfrac{1}{\sqrt{11}}\tan^{-1}\left(\dfrac{4\tan x + 1}{\sqrt{11}}\right) + c 11 1 tan − 1 ( 11 4 tan x + 1 ) + c Tap an option to check your answer.
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