# Tutor profile: Rose B.

## Questions

### Subject: Trigonometry

Verify the identity: $$\sin{x}\cdot \tan{x}=\sec{x}-\cos{x}$$

$$\sin{x}\cdot \tan{x}=\sec{x}-\cos{x}$$ $$\sin{x} \cdot \tan{x}=\frac{1}{\cos{x}}-\cos{x}$$ $$\sin{x} \cdot \tan{x}=\frac{1}{\cos{x}}-\frac{\cos^2{x}}{\cos{x}}$$ $$\sin{x} \cdot \tan{x}=\frac{1-\cos^2{x}}{\cos{x}}$$ $$\sin{x} \cdot \tan{x}=\frac{\sin^2{x}}{\cos{x}}$$ $$\sin{x} \cdot \tan{x}=\sin{x}\cdot \frac{\sin{x}}{\cos{x}}$$ $$\sin{x} \cdot \tan{x}=\sin{x}\cdot \tan{x}$$

### Subject: Calculus

Evaluate the definite integral: $$\int e^{2x}\cdot cos(x)dx$$

This problem is a great example of when to use multiple application of integration by parts. Keep in mind, integration by parts tell us that: $$\int u\cdot dv=u\cdot v-\int v\cdot du$$ For this application, it actually won't make a difference which term we use as our u and which we assign to dv. Let's look at the first application of integration by parts: $$u=e^{2x}\rightarrow du=2\cdot e^{2x}dx$$ $$dv=cos(x)dx\rightarrow v=sin(x)$$ $$\int e^{2x}\cdot cos(x)dx=e^{2x}\cdot sin(x)-\int 2 e^{2x}\cdot sin(x) dx$$ Doesn't seem much simpler, right? Well, let's apply integration by parts again, this time to the integral created in the first application. $$u=2e^{2x}\rightarrow du=4e^{2x}dx$$ $$dv=sin(x)dx\rightarrow v=-cos(x)$$ Putting this into our equation gives us: $$\int e^{2x}\cdot cos(x)dx=e^{2x}\cdot sin(x)-[-2e^{2x}\cdot cos(x)-\int -4e^{2x}\cdot cos(x)dx]$$ This looks messier, but it's actually close to done. Let's distribute all negatives and pull all constants out of our integral. $$\int e^{2x}\cdot cos(x)dx=e^{2x}\cdot sin(x)+2e^{2x}\cdot cos(x)-4\int e^{2x}\cdot cos(x)dx$$ What do you notice about the original integral we were trying to solve and the integral at the end of the expression? They are the same! Let's combine them on the left side. $$5\int e^{2x}\cdot cos(x)dx=e^{2x}\cdot sin(x)+2e^{2x}\cdot cos(x)$$ Now we simply divide by the coefficient and... $$\int e^{2x}\cdot cos(x)dx=\frac{e^{2x}\cdot sin(x)+2e^{2x}\cdot cos(x)}{5}+C$$ Voila! The solution!

### Subject: Algebra

Find all the solutions to the equation using factoring: $$x^4+4x^3-3x^2-12x=0$$

The first thing we need to do is determine if there are any common factors to the terms. We see that there is one: x. Factoring x out of every term gives us: $$x(x^3+4x^2-3x-12)=0$$ Now we need to determine how we will factor the polynomial that remains in the parentheses. Since this is a third-degree polynomial with four terms I would typically try factoring by grouping. This means we take the first two terms and factor out what they have in common and then take the third and fourth terms and factor what they have in common. If all goes according to plan, we should get the same term remaining from each group. $$x[x^2(x+4)-3(x+4)]=0$$ It worked! So now we can factor the common term from the inside, which is $$(x+4)$$. $$x(x-4)(x^2-3)=0$$ Now we simply apply the zero product property by taking each factored term and setting it equal to 0. $$ x=0$$ $$ x-4=0 \rightarrow x=4$$ $$x^2-3=0 \rightarrow x^2=3 \rightarrow x=\pm\sqrt{3}$$

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