Question

If $I_{m,n} = \int cos^m x sinnx \,dx$, then $7I_{4,3}-4I_{3,2}$ is equal to:

Answer 1

3 cos^5 x - 4 cos^7 x

Answer 2

Let the given integral be $I_{m,n} = \int \cos^m x \sin(nx) \,dx$. We need to evaluate $7I_{4,3}-4I_{3,2}$.

First, let's write out the terms: $I_{4,3} = \int \cos^4 x \sin(3x) \,dx$ $I_{3,2} = \int \cos^3 x \sin(2x) \,dx$

Substitute these into the expression: $7I_{4,3}-4I_{3,2} = 7 \int \cos^4 x \sin(3x) \,dx - 4 \int \cos^3 x \sin(2x) \,dx$ Combine the integrals: $7I_{4,3}-4I_{3,2} = \int (7 \cos^4 x \sin(3x) - 4 \cos^3 x \sin(2x)) \,dx$

Now, we use the trigonometric identity for $\sin(2x)$: $\sin(2x) = 2 \sin x \cos x$. Substitute this into the second term of the integrand: $4 \cos^3 x \sin(2x) = 4 \cos^3 x (2 \sin x \cos x) = 8 \cos^4 x \sin x$.

Substitute this back into the integral: $7I_{4,3}-4I_{3,2} = \int (7 \cos^4 x \sin(3x) - 8 \cos^4 x \sin x) \,dx$ Factor out $\cos^4 x$: $7I_{4,3}-4I_{3,2} = \int \cos^4 x (7 \sin(3x) - 8 \sin x) \,dx$

Now, use the triple angle identity for $\sin(3x)$: $\sin(3x) = 3 \sin x - 4 \sin^3 x$. Substitute this into the expression in the parenthesis: $7 \sin(3x) - 8 \sin x = 7(3 \sin x - 4 \sin^3 x) - 8 \sin x$ $= 21 \sin x - 28 \sin^3 x - 8 \sin x$ $= 13 \sin x - 28 \sin^3 x$.

Substitute this back into the integral: $7I_{4,3}-4I_{3,2} = \int \cos^4 x (13 \sin x - 28 \sin^3 x) \,dx$ Distribute $\cos^4 x$: $7I_{4,3}-4I_{3,2} = \int (13 \cos^4 x \sin x - 28 \cos^4 x \sin^3 x) \,dx$

Now, we can express $\sin^3 x$ in terms of $\sin x$ and $\cos x$: $\sin^3 x = \sin x \sin^2 x = \sin x (1 - \cos^2 x)$. Substitute this into the second term: $28 \cos^4 x \sin^3 x = 28 \cos^4 x \sin x (1 - \cos^2 x)$ $= 28 \cos^4 x \sin x - 28 \cos^6 x \sin x$.

Substitute this back into the integral: $7I_{4,3}-4I_{3,2} = \int (13 \cos^4 x \sin x - (28 \cos^4 x \sin x - 28 \cos^6 x \sin x)) \,dx$ $7I_{4,3}-4I_{3,2} = \int (13 \cos^4 x \sin x - 28 \cos^4 x \sin x + 28 \cos^6 x \sin x) \,dx$ $7I_{4,3}-4I_{3,2} = \int (-15 \cos^4 x \sin x + 28 \cos^6 x \sin x) \,dx$

Now, perform the integration. Let $u = \cos x$, then $du = -\sin x \,dx$, so $\sin x \,dx = -du$. The integral becomes: $\int (-15 u^4 (-du) + 28 u^6 (-du))$ $= \int (15 u^4 - 28 u^6) \,du$ Integrate term by term: $= 15 \frac{u^{4+1}}{4+1} - 28 \frac{u^{6+1}}{6+1} + C$ $= 15 \frac{u^5}{5} - 28 \frac{u^7}{7} + C$ $= 3 u^5 - 4 u^7 + C$

Substitute back $u = \cos x$: $7I_{4,3}-4I_{3,2} = 3 \cos^5 x - 4 \cos^7 x + C$.

The options provided are functions of $x$, implying the constant of integration is omitted or assumed to be zero.

The final answer is $3 \cos^5 x - 4 \cos^7 x$.