Q.(a) Sketch the output of the following circuit, where $V_i = 5\sin 100\pi t$.

The inverting amplifier first scales and inverts the sine, then the integrator integrates it; the integral of a sine is a negative cosine, so the output is a cosine of amplitude $7.5/\pi$ volts with a negative sign. …

[!TLDR]

Stage 1 gives $V_{o1}=-7.5\sin100\pi t$; the integrator ($1/R_iC=100$) turns this into $V_o=-2.38\cos100\pi t$ volts - a cosine wave of peak $2.38\,\text{V}$.

This II PUC Electronics op-amp problem chains an inverting amplifier into an integrator. Key idea: an ideal integrator obeys $V_o=-\frac{1}{R_iC}\int V_{in}\,dt$, and $\int\sin\omega t\,dt = -\frac{\cos\omega t}{\omega}$ - the minus sign is what makes an integrated sine come out as a cosine of opposite sign.

Stage 1 - Inverting amplifier:

Stage 2 - Integrator. Time constant $R_iC = 100\times10^3\times0.1\times10^{-6} = 0.01\,\text{s}$, so $1/R_iC = 100$:

Using $\int\sin100\pi t\,dt = -\dfrac{\cos100\pi t}{100\pi}$: …