Exams Joint Entrance Examination Mathematics Indefinite Integrals Question 4 Question 4 of 416 Medium MCQ
Let f ( t ) = ∫ ( 1 − sin ( log e t ) 1 − cos ( log e t ) ) d t , t > 1 f(t)=\int (\frac{1-\sin ({\log }_{e}t)}{1-\cos ({\log }_{e}t)})dt,t>1 f ( t ) = ∫ ( 1 − c o s ( l o g e t ) 1 − s i n ( l o g e t ) ) d t , t > 1 .
If f ( e π / 2 ) = − e π / 2 f({e}^{\pi /2})=-{e}^{\pi /2} f ( e π /2 ) = − e π /2 and f ( e π / 4 ) = α e π / 4 f({e}^{\pi /4})=\alpha {e}^{\pi /4} f ( e π /4 ) = α e π /4 , then α \alpha α equals
Solutions · 1 Post your solution Official solution
f ( t ) = ∫ ( 1 − sin ( log e t ) 1 − cos ( log e t ) ) d t , t > 1 f(t)=\int (\frac{1-\sin ({\log }_{e}t)}{1-\cos ({\log }_{e}t)})dt,t>1 f ( t ) = ∫ ( 1 − c o s ( l o g e t ) 1 − s i n ( l o g e t ) ) d t , t > 1
Let log e t = x {\log }_{e}t=x log e t = x .
t = e x ⇒ d t = e x d x f ( t ) = ∫ ( 1 − sin x 1 − cos x ) e x d x = ∫ ( 1 − 2 sin x 2 cos x 2 2 sin 2 x 2 ) e x d x sin 2 θ = 2 sin θ cos θ 1 − cos 2 θ = 2 sin 2 θ f ( t ) = ∫ [ 1 2 csc 2 x 2 − cot x 2 ] e x d x \begin{gathered} t={e}^{x}\Rightarrow dt={e}^{x}dx \\ f(t)=\int (\frac{1-\sin x}{1-\cos x}){e}^{x}dx \\ =\int (\frac{1-2\sin \frac{x}{2}\cos \frac{x}{2}}{2{\sin }^{2}\frac{x}{2}}){e}^{x}dx \begin{gathered} \sin 2\theta =2\sin \theta \cos \theta \\ 1-\cos 2\theta =2{\sin }^{2}\theta \end{gathered} \\ f(t)=\int [\frac{1}{2}{\csc }^{2}\frac{x}{2}-\cot \frac{x}{2}]{e}^{x}dx \end{gathered} t = e x ⇒ d t = e x d x f ( t ) = ∫ ( 1 − cos x 1 − sin x ) e x d x = ∫ ( 2 sin 2 2 x 1 − 2 sin 2 x cos 2 x ) e x d x sin 2 θ = 2 sin θ cos θ 1 − cos 2 θ = 2 sin 2 θ f ( t ) = ∫ [ 2 1 csc 2 2 x − cot 2 x ] e x d x
This standard integral of type ∫ e x [ g ( x ) + g ′ ( x ) ] = e x ⋅ g ( x ) + C \int {e}^{x}[g(x)+{g}^{'}(x)]={e}^{x}\cdot g(x)+C ∫ e x [ g ( x ) + g ′ ( x )] = e x ⋅ g ( x ) + C .
let g ( x ) = − cot x 2 g(x)=-\cot \frac{x}{2} g ( x ) = − cot 2 x .
g ′ ( x ) = − ( − csc 2 x 2 ) ⋅ 1 2 = 1 2 csc 2 x 2 . {g}^{'}(x)=-(-{\csc }^{2}\frac{x}{2})\cdot \frac{1}{2}=\frac{1}{2}{\csc }^{2}\frac{x}{2}. g ′ ( x ) = − ( − csc 2 2 x ) ⋅ 2 1 = 2 1 csc 2 2 x .
so f ( t ) = e x ⋅ ( − cot x 2 ) + C f(t)={e}^{x}\cdot (-\cot \frac{x}{2})+C f ( t ) = e x ⋅ ( − cot 2 x ) + C .
substitute value of x = log e t x={\log }_{e}t x = log e t back.
f ( t ) = − t cot ( log e t 2 ) + C f(t)=-t\cot (\frac{{\log }_{e}t}{2})+C f ( t ) = − t cot ( 2 l o g e t ) + C
use f ( e π / 2 ) = − e π / 2 f({e}^{\pi /2})=-{e}^{\pi /2} f ( e π /2 ) = − e π /2 to find C :
− e π / 2 = − e π / 2 cot ( log e e π / 2 2 ) + C − e π / 2 = − e π / 2 cot ( π 4 ) + C − e π / 2 = − e π / 2 + C ⇒ C = 0 \begin{gathered} -{e}^{\pi /2}=-{e}^{\pi /2}\cot (\frac{{\log }_{e}{e}^{\pi /2}}{2})+C \\ -{e}^{\pi /2}=-{e}^{\pi /2}\cot (\frac{\pi }{4})+C \\ -{e}^{\pi /2}=-{e}^{\pi /2}+C\Rightarrow C=0 \end{gathered} − e π /2 = − e π /2 cot ( 2 log e e π /2 ) + C − e π /2 = − e π /2 cot ( 4 π ) + C − e π /2 = − e π /2 + C ⇒ C = 0
f ( t ) = − t cot ( log e t 2 ) f(t)=-t\cot (\frac{{\log }_{e}t}{2}) f ( t ) = − t cot ( 2 l o g e t )
It is given that f ( e π / 4 ) = α e π / 4 f({e}^{\pi /4})=\alpha {e}^{\pi /4} f ( e π /4 ) = α e π /4 .
α e π / 4 = − e π / 4 cot ( log e e π / 4 2 ) α = − cot ( π 8 ) \begin{gathered} \alpha {e}^{\pi /4}=-{e}^{\pi /4}\cot (\frac{{\log }_{e}{e}^{\pi /4}}{2}) \\ \alpha =-\cot (\frac{\pi }{8}) \end{gathered} α e π /4 = − e π /4 cot ( 2 log e e π /4 ) α = − cot ( 8 π )
using result 1 + cos 2 θ = 2 cos 2 θ . . . ( 1 ) 1+\cos 2\theta =2{\cos }^{2}\theta \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ...(1) 1 + cos 2 θ = 2 cos 2 θ ... ( 1 )
sin 2 θ = 2 sin θ cos θ 1 + cos 2 θ sin 2 θ = 2 cos 2 θ 2 sin θ cos θ = cot θ . . . ( 2 ) \begin{gathered} \sin 2\theta =2\sin \theta \cos \theta \\ \frac{1+\cos 2\theta }{\sin 2\theta }=\frac{2{\cos }^{2}\theta }{2\sin \theta \cos \theta }=\cot \theta \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ...(2) \end{gathered} sin 2 θ = 2 sin θ cos θ sin 2 θ 1 + cos 2 θ = 2 sin θ cos θ 2 cos 2 θ = cot θ ... ( 2 )
α = − cot ( π 8 ) = − ( 1 + cos π 4 sin π 4 ) \alpha =-\cot (\frac{\pi }{8})=-(\frac{1+\cos \frac{\pi }{4}}{\sin \frac{\pi }{4}}) α = − cot ( 8 π ) = − ( s i n 4 π 1 + c o s 4 π )
α = − ( 1 + 1 2 1 2 ) = − ( 2 + 1 ) \alpha =-(\frac{1+\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}})=-(\sqrt{2}+1) α = − ( 2 1 1 + 2 1 ) = − ( 2 + 1 ) . Answer the question to unlock