Cho hàm số y = f ( x ) có f ( 0 ) = 0 và f ′ ( x ) = sin 8x − cos 8x − 4 sin 6x , ∀ x ∈ R . Tính I = pi ∫ 0 16 f ( x ) dx .

19/50

Cho hàm số \(y = f\left( x \right)\) có \(f\left( 0 \right) = 0\) và \(f'\left( x \right) = {\rm{si}}{{\rm{n}}^8}x - {\rm{co}}{{\rm{s}}^8}x - 4{\rm{si}}{{\rm{n}}^6}x,\forall x \in \mathbb{R}\). Tính I=∫0π16fxdx.

\(I = 10{\pi ^2}\).

\(I = 160\pi \).

\(I = 16{\pi ^2}\).

\(I = - 10{\pi ^2}\).

Giải thích

Ta có: \({\rm{si}}{{\rm{n}}^8}x - {\rm{co}}{{\rm{s}}^8}x - 4{\rm{si}}{{\rm{n}}^6}x = \left( {{\rm{si}}{{\rm{n}}^4}x - {\rm{co}}{{\rm{s}}^4}x} \right)\left( {{\rm{si}}{{\rm{n}}^4}x + {\rm{co}}{{\rm{s}}^4}x} \right) - 4{\rm{si}}{{\rm{n}}^6}x\)

\( = \left( {{\rm{si}}{{\rm{n}}^2}x - {\rm{co}}{{\rm{s}}^2}x} \right)\left( {{\rm{si}}{{\rm{n}}^4}x + {\rm{co}}{{\rm{s}}^4}x} \right) - 4{\rm{si}}{{\rm{n}}^6}x\)

\( = {\rm{co}}{{\rm{s}}^4}x{\rm{si}}{{\rm{n}}^2}x - {\rm{si}}{{\rm{n}}^4}x{\rm{co}}{{\rm{s}}^2}x - {\rm{co}}{{\rm{s}}^6}x - 3{\rm{si}}{{\rm{n}}^6}x\)

\( = {\rm{co}}{{\rm{s}}^4}x{\rm{si}}{{\rm{n}}^2}x - {\rm{si}}{{\rm{n}}^4}x{\rm{co}}{{\rm{s}}^2}x - 2{\rm{si}}{{\rm{n}}^6}x - \left( {{\rm{co}}{{\rm{s}}^6}x + {\rm{si}}{{\rm{n}}^6}x} \right)\)

\( = {\rm{si}}{{\rm{n}}^2}x\left( {{\rm{co}}{{\rm{s}}^4}x - {\rm{si}}{{\rm{n}}^4}x} \right) - {\rm{si}}{{\rm{n}}^4}x\left( {{\rm{co}}{{\rm{s}}^2}x + {\rm{si}}{{\rm{n}}^2}x} \right) - \left( {1 - 3{{\cos }^2}x{\rm{si}}{{\rm{n}}^2}x} \right)\)

\( = 4{\cos ^2}x{\rm{si}}{{\rm{n}}^2}x - 2{\rm{si}}{{\rm{n}}^4}x - 1 = - \frac{3}{4}{\rm{cos}}4x + {\rm{cos}}2x - \frac{5}{4}\).

Suy ra: fx=∫f'xdx=∫sin8x−cos8x−4sin6xdx

=∫−34cos4x+cos2x−54dx=−316sin4x+12sin2x−54x+C

Vì f0=0⇒C=0

Vậy fx=−316sin4x+12sin2x−54x

Suy ra I=∫0π16fxdx=∫0π16−316sin4x+12sin2x−54xdx

=∫0π−3sin4x+8sin2x−20xdx=34cos4x−4cos2x−10x20π=−10π2

Chọn D.