Evaluate the integrals in Exercises 41–60.
43. ∫6cosh(x/2 - ln3)dx

검증된 단계별 안내

Recall the definition of the hyperbolic cosine function: \(\cosh(u) = \frac{e^{u} + e^{-u}}{2}\). This can help in rewriting the integral if needed.

Identify the inner function inside the hyperbolic cosine: here, \(u = \frac{x}{2} - \ln 3\). This will be useful for substitution.

Use substitution by letting \(u = \frac{x}{2} - \ln 3\). Then, compute \(\frac{du}{dx} = \frac{1}{2}\), which implies \(dx = 2 \, du\).

Rewrite the integral in terms of \(u\): \(\int 6 \cosh\left(\frac{x}{2} - \ln 3\right) dx = \int 6 \cosh(u) \cdot 2 \, du = \int 12 \cosh(u) \, du\).

Integrate \(\cosh(u)\) with respect to \(u\): \(\int \cosh(u) \, du = \sinh(u) + C\). So, the integral becomes \(12 \sinh(u) + C\). Finally, substitute back \(u = \frac{x}{2} - \ln 3\) to express the answer in terms of \(x\).

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.

영상 길이:

도움이 되었나요?

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Hyperbolic Cosine Function (cosh)

The hyperbolic cosine function, cosh(x), is defined as (e^x + e^(-x))/2. It is an even function and appears frequently in calculus problems involving hyperbolic functions. Understanding its properties and derivatives is essential for integrating expressions involving cosh.

Integration of Composite Functions

When integrating functions like cosh(ax + b), it is important to use substitution or recognize the integral form. The integral of cosh(u) with respect to u is sinh(u), so adjusting for the inner function's derivative is necessary to find the correct antiderivative.

Logarithmic Constants in Integration

Constants such as ln(3) inside the argument of a function affect the integration as shifts but do not change the integral's form. Recognizing that ln(3) is a constant helps simplify the integral by treating it as a constant shift in the variable.

Evaluate the integrals in Exercises 41–60.43. ∫6cosh(x/2 - ln3)dx