Random-Problem-pi-4-pi-2-7sin-x-3cos-x-dx-By-getting-the-antiderivative-of-the-trigonometric-functions-sin-x-dx-cos-x-c-cos-x-dx-sin-x-c-7-sin-x-3-cos-x-pi-4-

Question Number 150037 by RoswelCod2003 last updated on 09/Aug/21

R a n d o m P r o b l e m :

\underset{\frac{π}{4}}{\int^{\frac{π}{2}}} (- 7 \sin x + 3 \cos x) d x

B y g e t t i n g t h e a n t i d e r i v a t i v e o f t h e t r i g o n o m e t r i c f u n c t i o n s :

\int \sin (x) d x = - \cos x + c

\int \cos (x) d x = \sin x + c

= - 7 \int \sin x + 3 \int \cos x \underset{\frac{π}{4}}{\overset{\frac{π}{2}}{∣}} = - 7 (- \cos x) + 3 (\sin x) \underset{\frac{π}{4}}{\overset{\frac{π}{2}}{∣}}

= 7 \cos x + 3 \sin x \underset{\frac{π}{4}}{\overset{\frac{π}{2}}{∣}}

E v a l u a t e i t t o t h e t o p a n d b o t t o m l i m i t o f i n t e g r a t i o n :

= (7 \cos \cdot \frac{π}{2} + 3 \sin \cdot \frac{π}{2}) - (7 \cos \cdot \frac{π}{4} + 3 \sin \cdot \frac{π}{4})

= [7 (0) + 3 (1)] - [7 (\frac{\sqrt{2}}{2}) + 3 (\frac{\sqrt{2}}{2})]

= 3 - \frac{7 \sqrt{2}}{2} - \frac{3 \sqrt{2}}{2}

= 3 - \frac{10 \sqrt{2}}{2} o r 3 - 5 \sqrt{2}

A n s w e r : 3 - 5 \sqrt{2}

S o l u t i o n b y R o s w e l :)