If \int \frac1a^2 \sin ^2 x+b^2 \cos ^2 x ~d x=\frac112 \tan ^-1(3 \tan x)+ constant, then the maximum value of a \sin x+b \cos x , is :

Indefinite Integrals PYQ — JEE Main 2024 · 6 Apr · Shift 2 · Q48 with Solution

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Correct answer: C

We use the standard substitution $t=\tan x$ .

Given $I=\int \frac{dx}{a^2\sin^2 x+b^2\cos^2 x}$ and it is equal to $\frac{1}{12}\tan^{-1}(3\tan x)+C.$

Rewrite the denominator in terms of $\tan x$ .

Using $\sin^2 x=\frac{\tan^2 x}{1+\tan^2 x},\qquad \cos^2 x=\frac{1}{1+\tan^2 x},$ we get $a^2\sin^2 x+b^2\cos^2 x=\frac{a^2\tan^2 x+b^2}{1+\tan^2 x}.$

Also, $dx=\frac{dt}{1+t^2},\qquad t=\tan x.$

Hence

=\int \frac{dt}{a^2 t^2+b^2}.$$ 3. Integrate the standard form. We know $$\int \frac{dt}{\alpha^2 t^2+\beta^2}=\frac{1}{\alpha\beta}\tan^{-1}\left(\frac{\alpha t}{\beta}\right)+C.$$ So here, $$I=\frac{1}{ab}\tan^{-1}\left(\frac{a}{b}\tan x\right)+C.$$ 4. Compare with the given result. Given $$I=\frac{1}{12}\tan^{-1}(3\tan x)+C.$$ Comparing, $$\frac{1}{ab}=\frac{1}{12}\quad\Rightarrow\quad ab=12$$ and $$\frac{a}{b}=3.$$ 5. Solve for $a$ and $b$. From $$a=3b,$$ substitute into $ab=12$: $$(3b)(b)=12\Rightarrow 3b^2=12\Rightarrow b^2=4\Rightarrow b=2.$$ Then $$a=6.$$ 6. Find the maximum value of $a\sin x+b\cos x$. Now $$a\sin x+b\cos x=6\sin x+2\cos x.$$ For an expression of the form $$A\sin x+B\cos x,$$ the maximum value is $$\sqrt{A^2+B^2}.$$ Therefore, $$\max(6\sin x+2\cos x)=\sqrt{6^2+2^2}=\sqrt{36+4}=\sqrt{40}.$$ 7. Match with options. $$\sqrt{40}$$ is option **C**.

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View written solutionFree

Correct answer: C

We use the standard substitution $t=\tan x$ .

Given $I=\int \frac{dx}{a^2\sin^2 x+b^2\cos^2 x}$ and it is equal to $\frac{1}{12}\tan^{-1}(3\tan x)+C.$

Rewrite the denominator in terms of $\tan x$ .

Using $\sin^2 x=\frac{\tan^2 x}{1+\tan^2 x},\qquad \cos^2 x=\frac{1}{1+\tan^2 x},$ we get $a^2\sin^2 x+b^2\cos^2 x=\frac{a^2\tan^2 x+b^2}{1+\tan^2 x}.$

Also, $dx=\frac{dt}{1+t^2},\qquad t=\tan x.$

Hence

=\int \frac{dt}{a^2 t^2+b^2}.$$ 3. Integrate the standard form. We know $$\int \frac{dt}{\alpha^2 t^2+\beta^2}=\frac{1}{\alpha\beta}\tan^{-1}\left(\frac{\alpha t}{\beta}\right)+C.$$ So here, $$I=\frac{1}{ab}\tan^{-1}\left(\frac{a}{b}\tan x\right)+C.$$ 4. Compare with the given result. Given $$I=\frac{1}{12}\tan^{-1}(3\tan x)+C.$$ Comparing, $$\frac{1}{ab}=\frac{1}{12}\quad\Rightarrow\quad ab=12$$ and $$\frac{a}{b}=3.$$ 5. Solve for $a$ and $b$. From $$a=3b,$$ substitute into $ab=12$: $$(3b)(b)=12\Rightarrow 3b^2=12\Rightarrow b^2=4\Rightarrow b=2.$$ Then $$a=6.$$ 6. Find the maximum value of $a\sin x+b\cos x$. Now $$a\sin x+b\cos x=6\sin x+2\cos x.$$ For an expression of the form $$A\sin x+B\cos x,$$ the maximum value is $$\sqrt{A^2+B^2}.$$ Therefore, $$\max(6\sin x+2\cos x)=\sqrt{6^2+2^2}=\sqrt{36+4}=\sqrt{40}.$$ 7. Match with options. $$\sqrt{40}$$ is option **C**.

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