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If \int \frac{d\theta }{\cos ^{2}\theta \left ( \tan 2\theta +\sec 2\theta \right )}=\lambda \tan \theta +2\log_{e}\left | f\left ( \theta \right ) \right |+C where C is a constant of integration, then the ordered pair \left ( \lambda ,f\left ( \theta \right ) \right ) is equal to: 

Option: 1

(-1,1-\tan \theta )
 


Option: 2

(-1,1+\tan \theta )


Option: 3

(1,1+\tan \theta )

 


Option: 4

(1,1-\tan \theta )


Answers (1)

best_answer

 

 

Integration as Reverse Process of Differentiation -

Rules of integration:

f(x) and g(x) are functions with antiderivatives ∫ f(x) and  ∫ g(x) dx. Then,

\\ {\text { (a) } \int \operatorname{kf}(x) d x=k \int f(x) d x \text { for any constant } k .} \\ {\text { (b) } \int(f(x)+g(x)) d x=\int f(x) d x+\int g(x) d x} \\ {\text { (c) } \int(f(x)-g(x)) d x=\int f(x) d x-\int g(x) d x}

\\\mathrm{Since,\;\;\frac{d}{dx}\left ( F(x) \right )=f(x)\;\;\Leftrightarrow \;\;\int f(x)dx=F(x)+c}

Based on this definition and various standard formulas (which we studied in Limit, Continuity and Differentiability) we obtained the following important integration formulae, 

\\\mathrm{1.\;\;\frac{d}{d x}\left({kx}\right)=k \Rightarrow \int kd x=kx+C, \text{ where k is a constant}}\\\\\mathrm{2.\;\;\frac{d}{d x}\left(\frac{x^{n+1}}{n+1}\right)=x^{n}, n \neq-1 \Rightarrow \int x^{n} d x=\frac{x^{n+1}}{n+1}+C, n \neq-1}\\\\\mathrm{3.\;\;\frac{d}{d x}(\log |x|)=\frac{1}{x} \Rightarrow \int \frac{1}{x} d x=\log |x|+C, \text { when } x \neq 0}\\\\\mathrm{4.\;\;\frac{d}{d x}\left(e^{x}\right)=e^{x} \Rightarrow \int e^{x} d x=e^{x}+C}\\\\\mathrm{5.\;\;\frac{d}{d x}\left(\frac{a^{x}}{\log _{e} a}\right)=a^{x}, a>0, a \neq 1\;\;\Rightarrow \int a^{x} d x=\frac{a^{x}}{\log _{e} a}+C}

 

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\\ \int \frac{\sec ^{2} \theta}{\frac{1+\tan ^{2} \theta}{1-\tan ^{2} \theta}+\frac{2 \tan \theta}{1-\tan ^{2} \theta}} d \theta \\ {=\int \frac{\sec ^{2} \theta\left(1-\tan ^{2} \theta\right)}{(1+\tan \theta)^{2}} d \theta}\\=\int \frac{\sec ^{2} \theta(1-\tan \theta)}{1+\tan \theta} d \theta\\\text{put }\tan\theta=t\Rightarrow \sec^2\theta d\theta=dt\\

\\ {=\int\left(\frac{1-t}{1+t}\right) d t=\int\left(-1+\frac{2}{1+t}\right) d t} \\ {=-t+2 \log (1+t)+C} \\ {=-\tan \theta+2 \log (1+\tan \theta)+C} \\ {\Rightarrow \quad \lambda=-1 \text { and } f(x)=1+\tan \theta}

Correct option 2

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Rishabh

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