Exams Joint Entrance Examination Mathematics Indefinite Integrals Question 414 Question 414 of 416 Medium MCQ
Let ∫ x 3 sin x d x = g ( x ) + C \int x^3 \sin x \mathrm{~d} x=g(x)+C ∫ x 3 sin x d x = g ( x ) + C , where C C C is the constant of integration. If 8 ( g ( π 2 ) + g ′ ( π 2 ) ) = α π 3 + β π 2 + γ , α , β , γ ∈ Z 8\left(g\left(\frac{\pi}{2}\right)+g^{\prime}\left(\frac{\pi}{2}\right)\right)=\alpha \pi^3+\beta \pi^2+\gamma, \alpha, \beta, \gamma \in Z 8 ( g ( 2 π ) + g ′ ( 2 π ) ) = α π 3 + β π 2 + γ , α , β , γ ∈ Z , then α + β − γ \alpha+\beta-\gamma α + β − γ equals :
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\[ ∫ x 3 sin x d x = − x 3 cos x + ∫ 3 x 2 cos x d x = − x 3 cos x + 3 x 2 sin x − ∫ 6 x sin x d x = − x 3 cos x + 3 x 2 sin x + 6 x cos x − 6 sin x + c \begin{aligned}
& \int x^3 \sin x d x=-x^3 \cos x+\int 3 x^2 \cos x d x \\
& =-x^3 \cos x+3 x^2 \sin x-\int 6 x \sin x d x \\
& =-x^3 \cos x+3 x^2 \sin x+6 x \cos x-6 \sin x+c
\end{aligned} ∫ x 3 sin x d x = − x 3 cos x + ∫ 3 x 2 cos x d x = − x 3 cos x + 3 x 2 sin x − ∫ 6 x sin x d x = − x 3 cos x + 3 x 2 sin x + 6 x cos x − 6 sin x + c \]
So g ( x ) = − x 3 cos x + 3 x 2 sin x + 6 x cos x − 6 sin x g(x)=-x^3 \cos x+3 x^2 \sin x+6 x \cos x-6 \sin x g ( x ) = − x 3 cos x + 3 x 2 sin x + 6 x cos x − 6 sin x
\[ g ( π 2 ) = 3 π 2 4 − 6 g ′ ( x ) = x 3 sin x g ′ ( π 2 ) = π 3 8 8 ( g ( π 2 ) + g ′ ( π 2 ) ) = π 3 + 6 π 2 − 48 \begin{aligned}
& g\left(\frac{\pi}{2}\right)=\frac{3 \pi^2}{4}-6 \\
& g^{\prime}(x)=x^3 \sin x \\
& g^{\prime}\left(\frac{\pi}{2}\right)=\frac{\pi^3}{8} \\
& 8\left(g\left(\frac{\pi}{2}\right)+g^{\prime}\left(\frac{\pi}{2}\right)\right)=\pi^3+6 \pi^2-48
\end{aligned} g ( 2 π ) = 4 3 π 2 − 6 g ′ ( x ) = x 3 sin x g ′ ( 2 π ) = 8 π 3 8 ( g ( 2 π ) + g ′ ( 2 π ) ) = π 3 + 6 π 2 − 48 \]
So α + β − γ = 55 \alpha+\beta-\gamma=55 α + β − γ = 55 Answer the question to unlock