BITSAT 2022
Exam BITSAT
Year 2022
Subject Mathematics
Chapter Indefinite Integration BITSAT 2022 · Mathematics · Indefinite Integration
Let f(x)=∫ x^2 d x (1+x^2)(1+ .1+x^2) . and f(0)=0 , then the value...
Let
f ( x ) = ∫ x 2 d x ( 1 + x 2 ) ( 1 + 1 + x 2 ) f(x)=\int \frac{x^2 d x}{\left(1+x^2\right)\left(1+\sqrt{\left.1+x^2\right)}\right.} f ( x ) = ∫ ( 1 + x 2 ) ( 1 + 1 + x 2 ) x 2 d x and
f ( 0 ) = 0 f(0)=0 f ( 0 ) = 0 , then the value of
f ( 1 ) f(1) f ( 1 ) be
Options A log ( 1 + 2 ) \log (1+\sqrt{2}) log ( 1 + 2 ) B log ( 1 + 2 ) − π 4 \log (1+\sqrt{2})-\frac{\pi}{4} log ( 1 + 2 ) − 4 π C log ( 1 + 2 ) + π 2 \log (1+\sqrt{2})+\frac{\pi}{2} log ( 1 + 2 ) + 2 π D None of these Correct answer B. log ( 1 + 2 ) − π 4 \log (1+\sqrt{2})-\frac{\pi}{4} log ( 1 + 2 ) − 4 π
Step-by-step solution
f ( x ) = ∫ x 2 d x ( 1 + x 2 ) ( 1 + 1 + x 2 ) f(x)=\int \frac{x^2 d x}{\left(1+x^2\right)\left(1+\sqrt{1+x^2}\right)} f ( x ) = ∫ ( 1 + x 2 ) ( 1 + 1 + x 2 ) x 2 d x Let
x = tan θ ⇒ d x = sec 2 θ d θ = ( 1 + x 2 ) d \thetaf ( x ) = ∫ x 2 d x ( 1 + x 2 ) ( 1 + 1 + x 2 ) = ∫ tan 2 θ sec 2 θ d θ sec 2 θ ( 1 + sec θ ) = ∫ tan 2 θ d θ 1 + sec θ = ∫ sin 2 θ d θ cos θ ( 1 + cos θ ) = ∫ 1 − cos 2 θ d θ cos θ ( 1 + cos θ ) = ∫ ( 1 − cos θ ) d θ cos θ = ∫ sec θ d θ − ∫ d θ = log ( x + 1 + x 2 ) − tan − 1 x + C ∴ f ( 0 ) = log ( 0 + 1 + 0 ) − tan − 1 ( 0 ) + C 0 = log 1 − 0 + C ⇒ C = 0 ∴ f ( 1 ) = log ( 1 + 1 + 1 2 ) − tan − 1 ( 1 ) = log ( 1 + 2 ) − π 4 x=\tan \theta\Rightarrow \quad d x=\sec ^2 \theta d \theta=\left(1+x^2\right) d \thetaf(x)=\int \frac{x^2 d x}{\left(1+x^2\right)\left(1+\sqrt{1+x^2}\right)}=\int \frac{\tan ^2 \theta \sec ^2 \theta d \theta}{\sec ^2 \theta(1+\sec \theta)}=\int \frac{\tan ^2 \theta d \theta}{1+\sec \theta}=\int \frac{\sin ^2 \theta d \theta}{\cos \theta(1+\cos \theta)}=\int \frac{1-\cos ^2 \theta d \theta}{\cos \theta(1+\cos \theta)}=\int \frac{(1-\cos \theta) d \theta}{\cos \theta}=\int \sec \theta d \theta-\int d \theta=\log \left(x+\sqrt{1+x^2}\right)-\tan ^{-1} x+C\therefore \quad f(0)=\log (0+\sqrt{1+0})-\tan ^{-1}(0)+C0=\log 1-0+C\Rightarrow \quad C=0\therefore \quad f(1)=\log \left(1+\sqrt{1+1^2}\right)-\tan ^{-1}(1)=\log (1+\sqrt{2})-\frac{\pi}{4} x = tan θ ⇒ d x = sec 2 θ d θ = ( 1 + x 2 ) d \thetaf ( x ) = ∫ ( 1 + x 2 ) ( 1 + 1 + x 2 ) x 2 d x = ∫ s e c 2 θ ( 1 + s e c θ ) t a n 2 θ s e c 2 θ d θ = ∫ 1 + s e c θ t a n 2 θ d θ = ∫ c o s θ ( 1 + c o s θ ) s i n 2 θ d θ = ∫ c o s θ ( 1 + c o s θ ) 1 − c o s 2 θ d θ = ∫ c o s θ ( 1 − c o s θ ) d θ = ∫ sec θ d θ − ∫ d θ = log ( x + 1 + x 2 ) − tan − 1 x + C ∴ f ( 0 ) = log ( 0 + 1 + 0 ) − tan − 1 ( 0 ) + C 0 = log 1 − 0 + C ⇒ C = 0 ∴ f ( 1 ) = log ( 1 + 1 + 1 2 ) − tan − 1 ( 1 ) = log ( 1 + 2 ) − 4 π
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