Question

I=\(\int\limits^b_a\left(x+\dfrac{\pi}{6}\right)\) dx theo m,n biết rằng:

\(\int\limits^a_b\left(sinx+cosx\right)\) dx=m ;\(\int\limits^b_a\left(sinx-cosx\right)dx\)

=n

Answer 1

Bạn xem lại xem có type thiếu đề không? \((x+\frac{\pi}{6})\) có sin hay cos, tan ở phía trước không?

Answer 2

Sin nha

Answer 3

\(\int\limits^a_b\left(sinx+cosx\right)dx=\left(sinx-cosx\right)|^a_b=sina-cosa-sinb+cosb=m\)

\(\int\limits^b_a\left(sinx-cosx\right)dx=\left(-cosx-sinx\right)|^b_a=-cosa-sina+cosb+sinb=n\)

\(\Rightarrow\left\{{}\begin{matrix}m+n=-2\left(cosa-cosb\right)\\m-n=2\left(sina-sinb\right)\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}cosa-cosb=-\dfrac{m+n}{2}\\sina-sinb=\dfrac{m-n}{2}\end{matrix}\right.\)

\(I=\int\limits^b_asin\left(x+\dfrac{\pi}{6}\right)dx=-cos\left(x+\dfrac{\pi}{6}\right)|^b_a=cos\left(a+\dfrac{\pi}{6}\right)-cos\left(b+\dfrac{\pi}{6}\right)\)

\(=cosa.cos\left(\dfrac{\pi}{6}\right)-sina.sin\left(\dfrac{\pi}{6}\right)-cosb.cos\left(\dfrac{\pi}{6}\right)+sinb.sin\left(\dfrac{\pi}{6}\right)\)

\(=\dfrac{\sqrt{3}}{2}\left(cosa-cosb\right)-\dfrac{1}{2}\left(sina-sinb\right)\)

\(=\dfrac{-\sqrt{3}}{4}\left(m+n\right)-\dfrac{1}{4}\left(m-n\right)\)

Answer 4

Ủa ngáo ngáo, nhìn ngược cận con tích phân sau

\(\int\limits^a_b\left(sinx+cosx\right)dx=\left(sinx-cosx\right)|^a_b=sina-cosa-sinb+cosb=m\)

\(\int\limits^b_a\left(sinx-cosx\right)dx=\left(-sinx-cosx\right)|^b_a=sina+cosa-sinb-cosb=n\)

\(\Rightarrow\left\{{}\begin{matrix}m+n=2\left(sina-sinb\right)\\m-n=-2\left(cosa-cosb\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}sina-sinb=\dfrac{m+n}{2}\\cosa-cosb=-\dfrac{m-n}{2}\end{matrix}\right.\)

\(I=\dfrac{\sqrt{3}}{2}\left(cosa-cosb\right)-\dfrac{1}{2}\left(sina-sinb\right)\)

\(=\dfrac{-\sqrt{3}}{4}\left(m-n\right)-\dfrac{1}{4}\left(m+n\right)\)