For the following function \( \frac{\mathrm{us}}{\mathrm{dt}} \), find the antiderivative \( s \) that satisfies the given condition. \[ \frac{\mathrm{ds}}{\mathrm{dt}}=2 \sin 2 \mathrm{t}-16 \cos 16 \mathrm{t} ; \mathrm{s}\left(\frac{\pi}{4}\right)=6 \] The antiderivative that satisfies the given condition is \( s(t)=\square \).

Find the antiderivative of \( 2*\sin(2*t)-16*\cos(16*t) \) Evaluate the integral by following steps: - step0: Evaluate: \(\int 2\sin\left(2t\right)-16\cos\left(16t\right) dt\) - step1: Use properties of integrals: \(\int 2\sin\left(2t\right) dt-\int 16\cos\left(16t\right) dt\) - step2: Evaluate the integral: \(-\cos\left(2t\right)-\sin\left(16t\right)\) - step3: Add the constant of integral C: \(-\cos\left(2t\right)-\sin\left(16t\right) + C, C \in \mathbb{R}\) Substitute \( t=\frac{\pi}{4} \) into the expression \( -\cos(2t)-\sin(16t) + C \). Evaluate the expression by following steps: - step0: Evaluate: \(-\cos\left(2t\right)-\sin\left(16t\right)+C\) - step1: Substitute: \(-\cos\left(2\times \frac{\pi }{4}\right)-\sin\left(16\times \frac{\pi }{4}\right)+C\) - step2: Reduce the numbers: \(-\cos\left(\frac{\pi }{2}\right)-\sin\left(16\times \frac{\pi }{4}\right)+C\) - step3: Reduce the numbers: \(-\cos\left(\frac{\pi }{2}\right)-\sin\left(4\pi \right)+C\) - step4: Calculate: \(-1\times 0-\sin\left(4\pi \right)+C\) - step5: Calculate: \(-1\times 0-0+C\) - step6: Multiply: \(0-0+C\) - step7: Remove 0: \(C\) The antiderivative of \( \frac{ds}{dt}=2 \sin 2t-16 \cos 16t \) is \( -\cos(2t)-\sin(16t) + C \), where \( C \) is an arbitrary constant. Substitute \( t=\frac{\pi}{4} \) into the antiderivative, we get \( C \). Therefore, the antiderivative that satisfies the given condition is \( s(t)=-\cos(2t)-\sin(16t) + C \), where \( C \) is an arbitrary constant.